help-39

[hallow remnant]

yeah, for the purposes of such problems since we are not having fractional exponents, we can assume that it is just $\frac{\pi}{3}$

[jolly parrot]

Arnavutköy

[stoic imp]

okey

[hallow remnant]

so $(1+\sqrt{3}i)=(2,\frac{\pi}{3})$

[jolly parrot]

Arnavutköy

[hallow remnant]

what is $(1+\sqrt{3}i)^7$ then equal to in terms of polar coordinates?

[jolly parrot]

Arnavutköy

[stoic imp]

,, (1 + \sqrt{3}i)^7 = 2^7 \left(\cos\left(\frac{7\pi}{3}\right) + i\sin\left(\frac{7\pi}{3}\right)\right)

[hallow remnant]

no no need to do it that way

use de moivres again

what is $(2,\frac{\pi}{3})$ to the power of $7$?

[jolly parrot]

Arnavutköy

[hallow remnant]

we simply raise the magnitude to the power

but we only multiply the argument/angle

so what do we get

[stoic imp]

I applied demoivre but maybe I messed up

[hallow remnant]

um yeah let me tell which mistakes you did

[pure rapids]

Bro is helping 5 ppl at once

[stoic imp]

idk, it should be 2^7

[pure rapids]

What a 🗿

[hallow remnant]

• You should raise $2$ to the power of $7$

[jolly parrot]

Arnavutköy

[hallow remnant]

• It was $\frac{\pi}{3}$, not $\frac{\pi}{4}$, so you should change the $4$s to $3$s

[jolly parrot]

Arnavutköy

[stoic imp]

my bad yeah

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[hallow remnant]

it makes me feel productive which makes me feel less guilty for procrastinating finals studying

[pure rapids]

Ah

[pearl pond]

@stoic imp Has your question been resolved?

[hallow remnant]

oh sorry i forgot

um we can simplify $\frac{7\pi}{3}$ to $\frac{\pi}{3}$ as it cycles $\pmod 2\pi$ right?

[jolly parrot]

Arnavutköy

[stoic imp]

how so?

[stoic imp]

you said to not use the +2kpi

[hallow remnant]

well we can simplify still in terms of $2\pi$

[jolly parrot]

Arnavutköy

[hallow remnant]

but we don't need to "keep" track of all possible values

[stoic imp]

okay

,, (1 + \sqrt{3}i)^7 = 2^7 \left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[hallow remnant]

okay

okay so it is $(128,\frac{\pi}{3})$.

[jolly parrot]

Arnavutköy

[stoic imp]

,calc 2^7

[jolly parrot]

Result:

128

[hallow remnant]

Do you know what $\bar{z}$ is in terms of polar coordinates if $z=(r,\theta)$

[jolly parrot]

Arnavutköy

[stoic imp]

in terms of magnitude the conjugate has same magnitude

[hallow remnant]

yes

[stoic imp]

in terms of argument idk

[hallow remnant]

here ill tell you the argument becomes negative

basically imagine this on the polar coordinate grid

we get the conjugate by flipping over the x-axis right?

[stoic imp]

is hard for me to imagine, but the conjugate of

a + bi is a -bi
and for a - bi is a + bi

the imaginary part is opposite sign

x axis is Real part and y axis is imaginary part

idk if we flip over x axis, idk

I would need a drawing

[hallow remnant]

yeah ill just tell you it is $(r,-\theta)$.

[jolly parrot]

Arnavutköy

[hallow remnant]

This is a common fact

Thus, $\bar{z}^2=(r^2,-2\theta)$ right?

[jolly parrot]

Arnavutköy

[stoic imp]

,w arctan(-sqrt(3))

[jolly parrot]

Results provided by WolframAlpha

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[stoic imp]

its because of this prolly

[hallow remnant]

related yes

[stoic imp]

yeh

[hallow remnant]

okay, so multiplying these together, we have that $(1+\sqrt{3}i)^7\bar{z}^2=(2^7r^2,\frac{\pi}{3}-2\theta)$.

[jolly parrot]

Arnavutköy

[hallow remnant]

Therefore, we have $(64r^3,3\theta)=(128r^2,\frac{\pi}{3}-2\theta)$.

[jolly parrot]

Arnavutköy

[hallow remnant]

Can you find $r$ and $\theta$.

[jolly parrot]

Arnavutköy

[stoic imp]

θ = pi/15

r = 2

[stoic imp]

π/3-2θ? how did that happen

[hallow remnant]

when we multiply complex numbers the argument gets added

here, remember this golden rule.

multiplying numbers multiplies the magnitude, and adds the arguments

[stoic imp]

good

[hallow remnant]

okay, so when we set the LHS and the RHS equal in terms of the polar coordinates, we got that $(64r^3,3\theta)=(128r^2,\frac{\pi}{3}-2\theta)$. Can you solve for $r$ and $\theta$?

[jolly parrot]

Arnavutköy

[stoic imp]

θ = pi/15

r = 2

[hallow remnant]

good

so $z=(2,\frac{\pi}{15})$.

[jolly parrot]

Arnavutköy

[hallow remnant]

can you write $z$ now as a complex number?

[jolly parrot]

Arnavutköy

[hallow remnant]

your answer will contain sin and cosine

[stoic imp]

,, z = 2\left(\cos\left(\frac{\pi}{15}\right) + i\sin\left(\frac{\pi}{15}\right)\right)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[hallow remnant]

there is actually a few other solutions

remember how i told you that you could ignore the $2\pi$?

[jolly parrot]

Arnavutköy

[hallow remnant]

well now when we are setting these as equal, it isn't exactly true that $3\theta=\frac{\pi}{3}-2\theta$.

[jolly parrot]

Arnavutköy

[hallow remnant]

It is possible that they are a multiple of $2\pi$ off

[jolly parrot]

Arnavutköy

[stoic imp]

what to do then

[hallow remnant]

So suppose that $3\theta=\frac{\pi}{3}-2\theta+2\pi k$

[jolly parrot]

Arnavutköy

[hallow remnant]

Then $5\theta=\frac{\pi}{3}+2\pi k$.

[jolly parrot]

Arnavutköy

[hallow remnant]

Therefore, we get that $\theta$ can be $\frac{n\pi}{15}$, where $n$ can be $1,7,13,19,$ or $25$

[jolly parrot]

Arnavutköy

[stoic imp]

how so?

[hallow remnant]

plug in $k=0,1,2,3,4$

[jolly parrot]

Arnavutköy

[hallow remnant]

after $k=5$, you see that it repeats

[jolly parrot]

Arnavutköy

[stoic imp]

what happened with pi/3

[hallow remnant]

the $\frac{\pi}{3}$ case was when $k=0$. In this case, we got $\frac{\pi}{15}$.

[jolly parrot]

Arnavutköy

[stoic imp]

ok I see

,, \theta = \frac{\pi + 6 \pi k}{15}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

k = 0 ==> π/15

k=1 ==> 7π/15

k=2 ==> 13π/15

k = 3 ==> 19π/15

so what?

,, z = 2\left(\cos\left(\frac{\pi + 6 \pi k}{15}\right) + i\sin\left(\frac{\pi + 6 \pi k}{15}\right)\right)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

[hallow remnant]

yes this is right

[dreamy marsh]

Number 4?

[stoic imp]

<@&286206848099549185>

[hallow remnant]

use a new help channel

[hallow remnant]

you can just close this one

[stoic imp]

is this all z?

[hallow remnant]

yes it is

[stoic imp]

xdd

okay thankiu

.solved

[pearl pond]

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[pearl pond]

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[balmy scaffold]

not gonna lie. How do you find the solution for the left one

[orchid creek]

x² + 6x + 9 = x² + 8x + 16

[balmy scaffold]

ohh

[orchid creek]

6x + 9 = 8x + 16

So x should be -7/2

[balmy scaffold]

so just 2x = -7

[orchid creek]

Why -5?

9-16 is -7

[balmy scaffold]

my bad

[orchid creek]

Yea no worries

[balmy scaffold]

-wait no

it has exactly 1 solution and that is -7/2

[orchid creek]

Yes

Bc the x² cancels out

Also if you put it in the equation, it becomes (-½)² = (½)²

[pearl pond]

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[severe wadi]

this is my worksheet and I have a question ig about the topic

[dusty flame]

converting coordinates?

[severe wadi]

so I was doing it and I understood the lesson, and I thought it was just asking me to put the normal equations into polar equations

[dusty flame]

aight

so

plug x = rcostheta and y = rsintheta

then simplify

[severe wadi]

yeah so I got number one correct, but then for number 2, I did rcos(theta) + rsin(theta)=0

[dusty flame]

u want r on oneside

[severe wadi]

is that correct?

cuz then y did they solve for an angle

[dusty flame]

that is correct

but

once u factor -> ${r(\cos \theta + \sin \theta) = 0}$

[jolly parrot]

k

[severe wadi]

ohh ok

[dusty flame]

good?

[pearl pond]

@severe wadi Has your question been resolved?

[pearl pond]

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[sturdy phoenix]

how does one analyze functions of the same "family"?

[sturdy phoenix]

polynomial functions with parameters

like idk

x^2 plus ax

[green dew]

what do you mean same family?

[sturdy phoenix]

x^2 plus 1x/x^2 plus 2x/x^2 plus 3x

get the idea?

[green dew]

not really

[sturdy phoenix]

the function is pretty much the same besides the parameter

which is in this case 1,2,3

so they will have similiarities

whihc are calculateable

I'm not sure how thouhg

[green dew]

ohhh

for example ax + b

it has some common properties between different values of a and b?

[sturdy phoenix]

yes

[green dew]

lile 3x + 1 and 5x - 7?

[sturdy phoenix]

yep

[green dew]

ok well i understood that

now by analyzing

theres no difference?

[sturdy phoenix]

with analyze I mean calculate all extreme points, concavity and stuff like that

[green dew]

oh

js follow the rules that you know

abt extreme points or concavity or all that

there isnt much to do other than applying your formulas when analyzing functions

[pearl pond]

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[pearl pond]

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[topaz pasture]

Hello!!! Im studying for my calc final rn...but I don't understand 13c. How does the answer key get the bounds [-2,-3] and why is it being subtracted?

[pearl pond]

@topaz pasture Has your question been resolved?

[valid peak]

the idea is that for y between -3 and -2 the region is bounded only by the curve y=(x+2)^2-3

and the area of the sub-region bounded by that curve and y=-2 is given by this integral

this other integral corresponds to the area bounded between y=-2 and the two curves in the statement of the problem

these two curves meet at y=6

[topaz pasture]

ooooooooohhhhhhhhh

bro u made my brain fire like mad

neurons are ocnnecting

ok

ooooooooo bc under -2 there is only the curve

there isnt the linear line

wow

math sometimes is wild

ty tho

have a good one 🙂

.close

[pearl pond]

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[topaz pasture]

.reopen

[pearl pond]

✅

[topaz pasture]

wait I still confused on one part

why does the other integral not include the other branch of the equation

like the problem got split into -2-(y+3)^1/2 and -2+(y+3)^1/2

only the second equation used?

because x=-3 and x=-2 is still under the curve -2-(y+3)^1/2

[pearl pond]

@topaz pasture Has your question been resolved?

[valid peak]

no? they're integrating the difference between both branches between y=-3 and y=-2

[topaz pasture]

yes but the curve -2-(y+3)^1/2 still applies past y=-2

shouldnt the branches be (6,0) and (0,-2) and (-2,-3)???

[valid peak]

hmmm yes I agree with this

[valid peak]

so maybe there's a typo here

[topaz pasture]

because for (0,-2) the curve -2-(y+3)^1/2 still applies in the eqaution right?

Am i tripping?

like it's making sense right?

thank you for ur help though...

I'll prolly keep this open for a little bit longer to see if anybody else thinks I'm doing something wrong

[valid peak]

nevermind, I checked with desmos and this seems correct

[topaz pasture]

could you explain just a little bit more

[valid peak]

the branch x=-2+(y+3)^1/2 is the red curve here, while the blue line is x=(y-4)/2

note that the y and x axis labels are swapped here

[valid peak]

comparing to this picture you could think of x=-2+(y+3)^1/2 as being the "right side" inverse branch since it has a bigger value of x

[topaz pasture]

wait let me explore a little bit as well

ooooooohhhhh ok i seee

because the equation x=-2-(y+3)^1/2 ends after y=-2

hmmmmmm thats smart

i see i see

tysm for ur help

have a good one\

.close

[pearl pond]

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[pearl pond]

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[knotty prism]

I'm reviewing my assignments for AP Calc and currently working on derivatives of implicit realtions.

[knotty prism]

I am confused on number 3 and 6

I wanted to check if number 4 is right also

[regal herald]

i cant quite read 4 properly

for these things you need to use the product rule, and chain rule

did you write y'(e^x)(1)=0?

[pearl pond]

@knotty prism Has your question been resolved?

[knotty prism]

@regal herald yes

[regal herald]

thats not quite right

remember, product rule

y'e^x+ye^x=0

y'+y=0

y'=-y

[knotty prism]

Oke that makes sense

How would I setup for number 3

Would I separate x and natural log y?

So I would take the derivate of x and then the derivative of natural log y

[pearl pond]

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[knotty prism]

I meant the product rule lol

.reopen

[pearl pond]

✅

[regal herald]

3 is product rule on x and ln y yeah

(fg)'=f'g+fg'

(xlny)'=(x')lny+x(lny)'

[pearl pond]

@knotty prism Has your question been resolved?

[knotty prism]

What would the derivat9ve of natural log y be?

[pearl pond]

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[stoic imp]

A circle is circumscribed about an equilateral triangle with side lengths of $6$ units each. What is the area of the circle, in square units? Express your answer in terms of $\pi$.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[obsidian plume]

sup

[stoic imp]

hi, can I get some help

[strong relic]

i think you draw it out

label some stuff

and use pythagoras

[obsidian plume]

yea its been a while since i did this tryna remember how

[stoic imp]

?

[strong relic]

let the radius be r

and the height of the equilateral triangle be r+h

you can see 2 right angled truangles

so use pythagoras

[stoic imp]

how is this right angled

[strong relic]

if you extend the h

you will see it is also a height of the triangle

[obsidian plume]

u got the answer for this?

i got one but i wanna see if im right

before i explain how i got it

[stoic imp]

I am still trying to see what trickery ashley made

[obsidian plume]

ashy... but ashley sounds great

[strong relic]

i got 2 sqrt 3

[obsidian plume]

for area?

[strong relic]

for radius

area would be

18pi

[obsidian plume]

i got 4cos30 for r

[strong relic]

wait no

[obsidian plume]

12pi

[strong relic]

12 pi

[strong relic]

yeah same

[obsidian plume]

alr i guess my way works

[strong relic]

oh trig works better

[obsidian plume]

ye sqrt3/2

[strong relic]

cos the angles are nice

[obsidian plume]

yup

[stoic imp]

can i get some help down here?

[obsidian plume]

oh

yea

im writing it out so u can see

didnt use trig here, but i can do it that way if u want

its just the matter of using pythag vs trig to find x and h

[obsidian plume]

^^^

lemme know if any part confuses you

[stoic imp]

,rotate

[obsidian plume]

fyi, sqrt(27) can be simplified to 3sqrt(3), which then makes r 2sqrt(3)

then once u square it it becomes 4*3 which is 12

[stoic imp]

what does it say

[obsidian plume]

r+x = h

i write x very weird

[stoic imp]

why is that

[obsidian plume]

well i just split it into three

the very middle point

to the corners

as its a triangle, it divides by 3

[stoic imp]

how so?

[obsidian plume]

its an equaliteral triangle

so all the "bases" are 6

and the "heights" are from the corners to the middle of the triangle

hence all 3 have the same area

which combines into the big triangle

[stoic imp]

x is height of the shaded sub triangle

[obsidian plume]

yes

[stoic imp]

I understand the green but then what?

down or right?

[obsidian plume]

down

i try to find height

from the triangle under the circle

using pythag

[obsidian plume]

yellow part helps you find height

[stoic imp]

how is yellow related with red

[obsidian plume]

if u draw a circle around yellow

it becomes red

i didnt write 6 beside the sides

but its the same thing

its so hard to explain online lmao

[stoic imp]

how did u figured the height?

[obsidian plume]

look to the right

[stoic imp]

ahh Pythagoras

[obsidian plume]

ye

[stoic imp]

ok I understand how you got x

,w pi * (sqrt(27) - sqrt(27)/3)^2

[jolly parrot]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

oki, thanks

[obsidian plume]

👍

also its better if u keep things exact

and try without calculators when possible

cause in uni exams

i dont have a calculator D:

[stoic imp]

yes in pi

[obsidian plume]

lmao i meant with sqrt27

[stoic imp]

you are Viktor, u got it

[obsidian plume]

silco my man

not viktor

[stoic imp]

☠️

[obsidian plume]

alr lets close this lol

[stoic imp]

.solved

[pearl pond]

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[unreal stream]

Why must x by a n-dimensional vector of variables? Or, why can we not multiply matrix a by constant numbers?

[pearl pond]

@unreal stream Has your question been resolved?

[pearl pond]

@unreal stream Has your question been resolved?

[pearl pond]

@unreal stream Has your question been resolved?

[plush bramble]

You can multiply matrices by constants

c * A is the matrix A with all its entries multiplied by c

Top of page 2
https://math.mit.edu/~gs/dela/dela_5-1.pdf

[pearl pond]

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[pearl pond]

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[sacred hollow]

Help with question 30 please

[sacred hollow]

Please ping

[main oxide]

,rotate

[main oxide]

since it gets cut apart at x=1, I would try to make sure it's continuous at that point and has the same slope at that point by making sure the values are the same for both functions there

[pearl pond]

@sacred hollow Has your question been resolved?

[sacred hollow]

Thx

.close

[pearl pond]

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[gleaming notch]

I want to confirm that the following statement from Google's AI is correct,

You're absolutely right. While the traditional row echelon form and Gaussian elimination methods primarily deal with real numbers, the concept of linear equations and systems can be extended to the complex plane.
In such cases, the coefficients and variables can be complex numbers. The techniques for solving these systems are similar, but the calculations involve complex arithmetic.
However, the fundamental principles of row reduction and the concept of pivot elements still apply. The pivot elements can be complex numbers, and the row operations would involve complex arithmetic.
It's an interesting extension of linear algebra, and it has applications in various fields, including signal processing and quantum mechanics.

[gleaming notch]

The 'You're absolutely right.' statements of it are kinda suspicious.

It's very submissive, this llm.

[hallow remnant]

this is true

[gleaming notch]

Thanks mate!

.close

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[barren owl]

Prove that (sinv)^2+(cosv)^2=1. Justify clearly

[pure rapids]

We can prove using pythagorean theorem and then extend to all reals

[barren owl]

How so

I have no idea tbf

[pure rapids]

Ok so assume 0 < v < 90

[barren owl]

Ok

[pure rapids]

Then we can construct the right angle triangle where one angle is v and then it definition of sin and cos

Btw did they set a bound for v

[barren owl]

Shouldn’t it be

0<V<=90

[barren owl]

No information other than that given

[pure rapids]

Can as well

[barren owl]

Oh ok

[pure rapids]

Have you learnt that sin and cos can be extended to all real numbers

[barren owl]

Dont think so

[pure rapids]

Then all we need to do is prove for 0 < v < 90

[barren owl]

Okay

[pure rapids]

Add in the equal signs somewhere ig

[pure rapids]

Ok in the right triangle what is the definition of sin

And cos

[barren owl]

Opposite kathet/hypotenuse

Adj/hypotenuse

[pure rapids]

Yes

(Opp / hyp)^2 + (adj / hyp)^2 = 1 is what we are gonna prove

[barren owl]

Yes

[pure rapids]

Right

Ok

So just multiply both sides by hyp^2

[barren owl]

Uh ok

What does that give us

[pure rapids]

...

[barren owl]

Oh

B

Mb

[pure rapids]

Try just multiplying it and seeing if it's familiar

Hey no need to think so much

We get opp^2 + adj^2 = hyp^2

Is this familiar?

[barren owl]

Yea

[pure rapids]

What is it

[barren owl]

Pythagoran theroem

[pure rapids]

We're done at this step alr 🙂

[pure rapids]

Yes

[barren owl]

Ah okay

Seems so complicated until u do it

[pure rapids]

🙂 there is a version to extend to all real numbers which is that hard

[barren owl]

I rly appreciate it

[barren owl]

Im not that far along yet haha

[pure rapids]

Np

[barren owl]

Cya

[pure rapids]

Is ok

Bye!

[barren owl]

.close

[pearl pond]

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[strong relic]

the unit circle

(sin x, cos x) is any point on the unit circle

[pure rapids]

Yep

[strong relic]

using pythagoras shows sin^2 x + cos^2 x is the radius squared

[pearl pond]

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[strong relic]

which is 1

[strong relic]

oops

[pure rapids]

💀

[strong relic]

,close

[pure rapids]

.close

[strong relic]

.close

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[sharp smelt]

Prove that the nth roots of unity form a cyclic subgroup $C^{\cross}$ of order n

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

Kind of not sure where to start

I have to show that the roots have a generator

that's it, right

[tropic saddle]

yes

[sharp smelt]

hmm, $z^n=1 \implies z = 1^{\frac{1}{n}}$. It then follows that $z^2 = 1^{\frac{2}{n}}$ is also a root for $1^{\frac{2}{n}}^2=1$ and so on. It thus forms a cyclic group.

[jolly parrot]

ƒ(Why am. I here)=I don't Know

hmm, $z^n=1 \implies z = 1^{\frac{1}{n}}$. It then follows that $z^2 =  1^{\frac{2}{n}}$ is also a root for  $1^{\frac{2}{n}}^2=1$ and so on. It thus forms a cyclic group.
```Compilation error:```! Double superscript.
l.1422 ...}$ is also a root for  $1^{\frac{2}{n}}^
                                                  2=1$ and so on. It thus fo...
I treat `x^1^2' essentially like `x^1{}^2'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./796939110815629322.aux)
 ***********
LaTeX2e <2023-11-01>
L3 programming layer <2023-11-01>```

[tropic saddle]

please dont write it like that

which number is 1^1/n supposed to be

the whole point of nth roots of unity is that there are n of them

not just one

[sharp smelt]

a root of unity

[tropic saddle]

which root of unity

[sharp smelt]

the first, nth root of unity

[tropic saddle]

what is the "first" root of unity

[sharp smelt]

From $1=z^n$, we get that $z = 1^{\frac{1}{n}}$. As $z^n$ is 1, this tells us that $1^{\frac{1}{n}}$ is a root of unity. we also find that $z^a = 1^\frac{a}{n}$ is a root of unity, for $a \in \Z$ as $ {1^{\frac{a}{n}}}^n = 1^a=1$. This tells us that $1^{\frac{1}{n}}$ is a generator and that the set of the roots of unity forma. cyclic group.

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[tropic saddle]

I just told you not to write 1^1/n

[sharp smelt]

Then what do I write

[tropic saddle]

well what do you mean

[storm hatch]

$\omega_{n}$ like a normal person

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

just define it in the beginning

[tropic saddle]

which complex number is that

we are working in C

[sharp smelt]

oh, like hat

$e^{\frac{{i \theta}}{n}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tropic saddle]

whats theta

[sharp smelt]

The angle the complex number makes with the x-axis

[tropic saddle]

I mean, yes

but which angle is that

theta/n is the angle btw

[sharp smelt]

j 2π/n where j ranges from 0 to n-1 and n is the root that we're determining

[tropic saddle]

but which of this gives the "first" root

[sharp smelt]

j=0

[tropic saddle]

so e^0 = 1

?

[sharp smelt]

yes

[tropic saddle]

ok

so all nth roots are powers of 1?

[sharp smelt]

and then we move anti- clockwise

[sharp smelt]

hmm, no

they are powers of $e^{ \frac{2 i \pi}{n}}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[tropic saddle]

ok

so we set $\omega_n = e^{\frac{2 i\pi}{n}}$ and you are claiming that $\omega_n$ generates the group of $n$-th roots

[jolly parrot]

Denascite

[tropic saddle]

why

[sharp smelt]

yes

[sharp smelt]

wdym

[tropic saddle]

you are claiming that all nth roots are a power of omega_n

I am asking why

[sharp smelt]

$w_n=z$, so $w^n=z^n=1$. Similarly, $w^{2n}_n=z^{2n}=1^2=1, w^{an}_n=1^a=1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

does this work?

[tropic saddle]

that shows that powers of omega_n are also nth roots

it does not show that every nth root is a power of omega_n

[sharp smelt]

how do I do that

I mean I can get n roots this way, and by the fundamental theorm of algebra , there are n distinct roots

[sharp smelt]

Can I have a hint

[tropic saddle]

yes

exactly that

(well you dont need the full fta, you only need that there are <=n roots)

[sharp smelt]

so that justifies that this is a cyclic sub group

now I have to prove it's of order n

[tropic saddle]

that part should be obvious

[sharp smelt]

yeah, it kind of is

It's of order n as e^{2 iπ}=1

and not true for any integer less than n

thanks

.close

[pearl pond]

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[pearl pond]

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[runic rampart]

the blocked out part is when x=e

can someone explain how they did part b

[strong relic]

horizontal tangents occur when the derivtive is 0

[runic rampart]

but isnt the deriv x^1/x(1-lnx/x^2)

[strong relic]

yes

[runic rampart]

how did he only use 1-lnx

[strong relic]

set that to 0

and solve for x

because the other term

x^1/x-2

can never be 0

[runic rampart]

oh

[runic rampart]

can you explain this one too

why did they use ln

shouldnt the deriv be 2.847 t(1.0072)^t-1

actually i do remember this being before we learned the power rule

i forgot how u get the derivs without that

[versed remnant]

$$\frac{d}{dx} a^x = a^x \ln a$$ where a is any constant

[jolly parrot]

JustToPro

[runic rampart]

Oh I forgot about that

Thanks

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[stiff mauve]

how do they take the derivative here

[storm hatch]

differentiating a sum is the same as summing a ton of derivatives

so you can just $\frac{d}{dt}$ the inside and $pq^{n-1}$ is obviously a constant in the $t$ world

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

if you do it $r$ times, then first of all, obviously your $t^{n}$ has become $t^{n-r}$ due to repeated use of the power rule. And secondly, you're gonna end up with a bunch of trailing coefficients on the $t^{n-r}$ term because everytime you differentiate you take the exponent and put it in front. That long train of coefficients looks like $n(n-1)(n-2)(n-3)\cdots(n-r+1)$ which is equivalent to $\frac{n!}{(n-r)!}$

[jolly parrot]

如月あやみ　Kisaragi Ayami

[stiff mauve]

everything makes sense to me right now except how n(n-1)(n-2)......(n-r+1) is equal to that permutation could u maybe elaborate on that?

[storm hatch]

$n(n-1)(n-2)\cdots(n-r+1)=n(n-1)(n-2)\cdots(n-r+1)\times\frac{(n-r)(n-r-1)(n-r-2)\cdots(2)(1)}{(n-r)(n-r-1)(n-r-2)\cdots(2)(1)}$

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

$n(n-1)(n-2)\cdots(n-r+1)=\textcolor{red}{n(n-1)(n-2)\cdots(n-r+1)}\times\frac{\textcolor{red}{(n-r)(n-r-1)(n-r-2)\cdots(2)(1)}}{\textcolor{green}{(n-r)(n-r-1)(n-r-2)\cdots(2)(1)}}$

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

the red parts form $n!$ and the green parts form $(n-r)!$

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

you could see this more easily with a small example

$7\times6\times5\times4=7\times6\times5\times4\times\frac{3\times2\times1}{3\times2\times1}=\frac{7\times6\times5\times4\times3\times2\times1}{3\times2\times1}=\frac{7!}{3!}$

[jolly parrot]

如月あやみ　Kisaragi Ayami

[storm hatch]

makes sense?

[stiff mauve]

HOW did u come up with this man

[stiff mauve]

yes thank you so much

[storm hatch]

nice

[pearl pond]

@stiff mauve Has your question been resolved?

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[hoary nacelle]

Here’s the question and it’s a really tricky integral question

[hoary nacelle]

Here’s my solution please verify it for me and it took me 3 hours for the solution

@rough forge

This one ChatGPT won’t even be able to give hint (I never used ChatGPT to solve questions for me but generally speaking I do ask directional questions like how to approach certain questions) but this one it won’t get me a hint 😭

[plush bramble]

Do you get the same answer if you find the area of the full petal and multiply by 6 instead of half a petal and multiply by 12

,w solve cos(3x/2)=0

[jolly parrot]

Results provided by WolframAlpha

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

Requested by rie.mann

[plush bramble]

Take eg. n=0,-1 and get two values for theta

[hoary nacelle]

I just got frustrated yesterday, and I decided to do small analysis on each blooms so I slept after setting up the boundary and I computed the integral today

Is the answer correct though and how to find the region in general?

[hoary nacelle]

The tricky part is if I do integral over [pi/3, 2pi/3] which seemed to be a full blown, the area for the petal will be omitted which is bizarre right?

This is almost tricky for trickiness

Only when you do integration at the exact range of half piece you get the area of half of the blooms and its parametrized so it’s hard to find where the other half of bloom is

The question you can just restrict it easily the boundaries though maybe you can rotate the curve but it will either way take long time, can’t believe the prof thinks of anything like this

<@&286206848099549185> can you guys help me with verifying the sol and is there easy way doing such integral?

<@&286206848099549185>

[cinder thistle]

hm

so my method would be to integrate from theta = 0 to theta = pi/4 and multiply by 12

so 6* integral from 0 to pi/4 cos^2(3/2x)x dx

so this

which is

[hoary nacelle]

Why though? I wonder the reasoning and can you teach me the method? Because I will still need to justify my reason though

I just want to know every single method for tackling this type of integral

[cinder thistle]

basically i find the area of a half petal

multiply by the number of half petals

[hoary nacelle]

The shaded petals or unshaded?

[cinder thistle]

unshaded

[hoary nacelle]

But why 0 pi/4 will have half of unshaded petal please teach me 😭😭

[cinder thistle]

sry

i got the wrong answer

should have put sin

sry

[hoary nacelle]

That’s genius !!!

[hoary nacelle]

Then i literally calculated it wrong again😩😩😩

I will redo it now

[cinder thistle]

this is

[hoary nacelle]

That’s brilliant you saved my life 🥰🥰🥰🥰

[cinder thistle]

thanks

[hoary nacelle]

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[honest drift]

i have to find this limit n->infinite

[honest drift]

the answer should be -1

but i get this:

wait sorry

what i get is 0

but the answer should be -1

yes

this is indeterminable inf-inf

but this is the algorithm that we learned

[static helm]

shouldn’t these be multiplied together?

[honest drift]

no

this is something over 1

[static helm]

yes

[honest drift]

and down is radical a + radical b (a and b being what i have)

[static helm]

oh

[honest drift]

square root not radical*

[static helm]

okay then

so what is sqrta + sqrtb

[honest drift]

those are these square roots but it was too long to write:

also with + not -

so i can reduce the square roots

[static helm]

okay i see

[honest drift]

like here

[static helm]

how are you getting ones here and where did the n^2 go

[honest drift]

mmm

what n^2?

[honest drift]

waittttt

this might be it

you are right

[static helm]

yippee

[honest drift]

let me fix it and ill send a picture

[static helm]

alright

[static helm]

nicely done

[honest drift]

thanks for your help

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[silent bramble]

here, can't you just set the index to 0 to infinity, since n=0 is just summing up 0?

so then you would have something like this

[merry carbon]

e.g. if you set x = 1 in the series, you’d effectively be dividing by zero (from the (x-1)^{n - 2} etc) and of course that’s very bad

[silent bramble]

oh I see, so you can only change the index like that when it is a power series where its just x^n, xo=0

but they did that here in this example?

oh wait i meant to send this earlier, the wifis not too good

[merry carbon]

For this one, because you’re multiplying by “an appropriate power of x”, it doesn’t really matter as much cause you won’t end up with negative indices after simplifying it

[silent bramble]

i see

but generally if you have the form (x-xo)^n, where xo is not 0, then you would have to keep indices the same (or sub t=x-1) or smth like that

to solve

wait but theres an error here??

[merry carbon]

They reindexed to have the series start at n = 1 instead, then noticed that you don’t get any dividing by zero issues, and that n = 0 gives you an “output” of zero (hence their comment above that bottom line)

[silent bramble]

ahh i dont like this reindexing stuff, its making me confused im gonna try another method, ill send a picture in a bit

before I keep going, can I do smth like this?

@merry carbon

[merry carbon]

You can do it like that and remove the “lower” terms sure

[silent bramble]

ok!

so then I have to say for n>=2

i would get the same answer?

im gonna continue and try it out, thank you!!

[merry carbon]

It should be at least equivalent, I would think

[silent bramble]

yup it works!!!

🎉yay

thank you

[pearl pond]

@silent bramble Has your question been resolved?

[pearl pond]

@silent bramble Has your question been resolved?

[silent bramble]

summary sheet!

[silent bramble]

ok byeee

gonna go do my exam💀

.close

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[midnight haven]

For clarity this is not a real quiz 💀 its a practice for my finals but um how do I use the vertical line test 😭

[midnight haven]

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[knotty bough]

can you anser this?
σ (n) ≤ Hn +ln (Hn)eHn

Where n is a positive integer
Hn is the n-th harmonic number
σ(n) is the sum of the positive integers divisible by n
For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?

[autumn ravine]

$\rho(n) \leq H(n)+ln(H(n))*e^{H(n)}$?

[jolly parrot]

Kai The Cat

[knotty bough]

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[autumn ravine]

what

[pearl pond]

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[native iron]

HI PLZ ZHEJLP

[native iron]

WAIT

OOPS

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[sharp smelt]

Prove that every subgroup of a cylic group is cyclic. Do this by working with exponents and use the description of the subgroups of $\Z^{+}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

Why does the last part make a difference

[valid peak]

reads more like a hint to me

[sharp smelt]

Firstly, as we know this is a sub-group(H), we have the identity element say $1$ in the subgroup, it has inverses so if $a \in H$, then $a^{-1} \in H$ . Lastly, we know if $a \in H, a^n \in H$, then $a^{n+1} \in H$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

As all the elements are of the form $a^m, m \in \Z$, the group is cyclic

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[valid peak]

maybe what I would do is consider the morphism $\bZ\to \langle a\rangle$, $m\mapsto a^m$ and work from there

[jolly parrot]

derivada.schwarziana

[sharp smelt]

I haven't done morphisms yet

[valid peak]

oh alright

[sharp smelt]

I suppose I just use the defn of a cylic group then

[valid peak]

yet you know that the subgroups of $\bZ$ under addition are $n\bZ$ for some $n$ right

[jolly parrot]

derivada.schwarziana

[sharp smelt]

yes

[valid peak]

the idea is that you want to look at the n dividing the order of the group <a>

[sharp smelt]

I suppose a proof by contrapositive would work particularly well here

[valid peak]

er in the finite case anyway

(an infinite cyclic group is isomorphic to (Z, +) though)

[sharp smelt]

hmm

yeah

but what about contrapositive, that would work well, right

if the sub group is not cyclic, then the group is not cyclic

[valid peak]

that might work yes

[sharp smelt]

If the subgroup is not cylic, there is atleast one element in it that's not a power of a generator, thus, the group isn;t cyclic either as there is one element that isn't an power of a genator either

[valid peak]

right, more precisely if some subgp. H isn't cyclic there exist some a,b in H such that a is not a power of b

I think this is enough?

[sharp smelt]

Thanks

.close

[pearl pond]

Channel closed

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[brave sluice]

i'm not so sure

integers mod 4 under addition is cyclic but 1 isn't a power of 2

[valid peak]

that's right, I wasn't so sure of the claim tbh

[rustic gate]

what is that example supposed to say exactly

i guess you phrased it badly

[sharp smelt]

I guess this works under multiplication?

.reopen

[pearl pond]

✅

[brave sluice]

what are the elements under multiplication?

[sharp smelt]

a^{-1},1.a, a^2 \dots

[brave sluice]

i mean

what is integers mod 4 under multiplication?

1 is an identity

0 doesn't have an inverse

[sharp smelt]

0,1,2,3

[brave sluice]

so 0 can't be in it

[sharp smelt]

so 1,2,3,4

[brave sluice]

4 is 0

[sharp smelt]

yes, I know

[brave sluice]

well 4 doesn't have an inverse either

[sharp smelt]

hmm

then it isn't a group, is it

[brave sluice]

no

[sharp smelt]

under multiplication atleast

it is under addition

[brave sluice]

yes

[sharp smelt]

So my proof covers groups under multiplication

[brave sluice]

uh

[sharp smelt]

I now have to cover groups under addition?

[brave sluice]

you have to cover groups under any possible operation

[sharp smelt]

so a contrapositive proof won't work

I mean I just have to show there is one element that isn't the power of a given generator

right

[brave sluice]

i'm not sure how to do it

i'm thinking of saying something like
G is generated by some element, call it a
then all elements of H are expressable as a^n for some n

i want to take the gcd of these n's

or something like that

[sharp smelt]

why is what I've done wrong

[brave sluice]

this seems sus

for example

uh

how do you know it's not the power of a generator in G that wasn't in H?

[sharp smelt]

wdym

[brave sluice]

oh

i had an idea

but uh

[brave sluice]

see when they tried to make your statement more rigorous it fell apart

[sharp smelt]

how has it fallen apart

[rustic gate]

H isn't cyclic if there does not exist an element g in H such that every element of H is a power of g

[brave sluice]

if some subgp. H isn't cyclic there exist some a,b in H such that a is not a power of b
this fails to imply G isn't cyclic

[sharp smelt]

yes

[brave sluice]

ye i had an idea

[sharp smelt]

hmm?

[brave sluice]

H has elements of the form a^n

the element with the smallest such n should be a generator of H i think?

uh but you'd have to reduce the n's mod the order of G

[rustic gate]

doesn't matter

pick an element of H which can be expressed as a^n for the least possible n

[sharp smelt]

Okay

[brave sluice]

i still feel a bit stuck

[sharp smelt]

same

[rustic gate]

pick h in H

it's in G so h = a^m for some m

express h as a power of a^n

[brave sluice]

how do you know you can though

[sharp smelt]

a^m = (a^n)k \implies k = m/n

[rustic gate]

suppose not

then m does not divide n

[brave sluice]

n does not divide m

[rustic gate]

yeah that's what i meant

too many letters man

m = qn + r, r < n

a^m = a^qn a^r

but a^qn is in H since a^n is

so a^-qn is in H

so a^r is

but r < n

contradiction

[brave sluice]

yes

[rustic gate]

n was assumed to be minimal

[brave sluice]

i like it

[sharp smelt]

hmm

makes sense

Thanks

Got it , can I close this now

wait

I'm still not sure I get why my method is wrong

If I make an. argument about the group, it should be good, right

[brave sluice]

because your arguments are always good? heheh

[sharp smelt]

[brave sluice]

what do u mean

[sharp smelt]

I mean If I add an argument that connects the subgroup to the group, my argument should work

right

[brave sluice]

yeah if you can show H not cyclic implies G not cyclic, that would suffice

[sharp smelt]

If G is not cylic, there's atleast one element in it that is not a power of the generator from G, but this element is also in G, so G isn't cyclic

[brave sluice]

u started with G, you meant H right?

[sharp smelt]

yes

[brave sluice]

idk

[sharp smelt]

sorry for pestering you with my solution

[brave sluice]

it's fine, i enjoy it

[sharp smelt]

I'll close this now

.close

[pearl pond]

Channel closed

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[pearl pond]

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[ivory shadow]

help please

[ivory shadow]

how to punch this on calculator to have squareroot of 3/2

[plush bramble]

Use $\frac{a}{b}\cdot\frac{c}{d}= \frac{ac}{bd}$

[ivory shadow]

how?

[jolly parrot]

riemann

[plush bramble]

Oh it's a special angle

Use unit circle

,tex .unit circle

[jolly parrot]

riemann

[plush bramble]

Inverse sine and cosine don't have the domain you're looking for.

You sometimes have to add some angle to the inverse trig value to get the right angle

[ivory shadow]

oh

i get it

thank you

.close

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[pearl pond]

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[feral citrus]

I'm just lost like do I have to do 1/3(2x-5)

[hallow remnant]

are C and D supposed to be the midpoints?

[feral citrus]

yea

[hallow remnant]

so PC and QD are the medians

[feral citrus]

Yes

[hallow remnant]

so G is the median right?

*centroid sorry

[feral citrus]

yes the centroid

[hallow remnant]

ok, i will tell you that there is a theorem

that the centroid lies 2/3rds the way along all medians

if you want to prove this fact

you can prove it using coordinate geometry

using the fact that the centroid is the average of all coordinates

so essentialy this means that $\overline{QG}=2\overline{GD}$

[jolly parrot]

Arnavutköy

[feral citrus]

why would you do 2 (GD) and not 2/3 (GD)?

[hallow remnant]

it would be $\overline{QG}=\frac{2}{3}\overline{QD}$

[jolly parrot]

Arnavutköy

[feral citrus]

thank you very much

.close

[pearl pond]

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[agile ridge]

Can anyone explain this step?

[compact ridge]

oh they just rewrote $\cos^3 = \cos \cos^2 = \cos (1 - \sin^2)$

[jolly parrot]

south

[compact ridge]

and then you have a -1 * -1 * 1/(sin^2 theta) left to integrate

[versed mica]

1-sin^2 = cos^2, so the first integral was just rewriting cos^2 and then they brought the -1 factor outside the integral so -1/3 -> 1/3 because now it’s + 1 instead of -1

[compact ridge]

there is a minus sign in front of the entire integral

[agile ridge]

Ooh thanks Is this a common trick😢 cuz it's really hard to come up with

[pearl pond]

@agile ridge Has your question been resolved?

[pearl pond]

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[wary heron]

Why does this go to -1/2

Shouldnt it go to 0

Limit is towards +inf btw

Of x^2-x(sqrt(x^2+1))

I multiply by conjugate

[dapper kraken]

you got $\lim_{x\to\infty}-\frac{1}{1+\sqrt{1+\frac{1}{x^2}}}$ from what i can see?

[jolly parrot]

Skissue ping4response

[pearl pond]

@wary heron Has your question been resolved?

[pearl pond]

Channel closed

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[wary heron]

.reopen

[pearl pond]

✅

[wary heron]

Ye

[dapper kraken]

when it goes to infinity, 1/x^2 goes to 0

[wary heron]

Gives -1/2 in total

Should be 0

Nvm im right

Lel

[pearl pond]

@wary heron Has your question been resolved?

[wary heron]

Maybe

[plucky python]

yh the answer is -1/2

[wary heron]

Maybe

[dapper kraken]

!done