help-49

[fallow scarab]

$a^x b^x = (ab)^x$ yes

[grand pond]

riemann

[violet dune]

okay

so now i have $\frac{162*6^x}{\ln{6}} + C$. is there anything i can do after that?

[grand pond]

Krish

[violet dune]

i just checked the answer key it says $\frac{27*6^{x+1}}{\ln{6}} + C$, which is the same thing. which would be considered more simplified? or does it matter which one i put as the answer? keep in mind this is practice for an integration bee i am participating in

[grand pond]

Krish

[tough shale]

It really depends on how you define 'simplified'

If u prefer only x's for exponents, then yours is right

If u prefer all constants to be sucked up by the exponent, then the book one is more simplified

[violet dune]

yeah true. what do you think they would ask for in an integration bee?

[tough shale]

I'm not sure about what is the standard for mathematics though

[violet dune]

or do you think they would care?

because imo theyre the same level of simplification

[tough shale]

I don't think they would care. Both of them give the same answer after differentiating again

I too think so

[violet dune]

thank you, and thanks to the other ppl for the help

i do have one more that im a little stuck on

$\int_1^2{(x^{x^2+1})(2\ln{x}+1)}dx$

[grand pond]

Krish

[violet dune]

i tried using integration by parts but how would i take the derivative of x^x^2+1

[tough shale]

That 2 beside lnx

That itches me to put there an x in between the 2 and lnx

[violet dune]

haha yeah 2xlnx looks a lot better

[tough shale]

And there is a x^1 present with the x^(x²+1)

Just use that and you'll see magic

[violet dune]

$x*x^{x^2}$

[grand pond]

Krish

[tough shale]

Yes

[violet dune]

so if i use product rule

that would be $x^{x^2} + x*$?

[grand pond]

Krish

[violet dune]

would it be 2x*x^x^2?

[tough shale]

You know how to differentiate x^x²?

[violet dune]

not really because theyre both functions of x

im used to doing derivatives of x^2 or 2^x, etc. but x^x is weird

[tough shale]

Let me teach you...

let $y = x^x$

[grand pond]

Suika

[tough shale]

We want $\frac{dy}{dx}$

[grand pond]

Suika

[violet dune]

yes

[tough shale]

Take ln on both the sides here

$\ln{y}=x\ln{x}$

[violet dune]

lny = xlnx?

[tough shale]

Mb

[grand pond]

Suika

[tough shale]

Yes

[violet dune]

ok cool yeah

[tough shale]

Now you can differentiate it

$\frac{1}{y}\cdot y'=\ln{x} + \frac{x}{x}$

[grand pond]

Suika

[violet dune]

$\frac{1}{y} = \ln{x} + 1 + C$?

[grand pond]

Krish

[tough shale]

There's the chain rule on lhs

[violet dune]

oh y'

yeah

[tough shale]

Yes

So $y'=x^x(\ln{x}+1)$

[grand pond]

Suika

[tough shale]

And there's a trick...

[violet dune]

so you just substituted y back in for x^x right?

[tough shale]

to do it quicker

Yes

[violet dune]

<@&268886789983436800>

[tough shale]

For $x^x$, if the exponent were constant, you would do $x\cdot x^{x-1}$ right?

[violet dune]

yeah

[grand pond]

Suika

[tough shale]

If the base were constant, you would do $x^x\ln{x}$

[grand pond]

Suika

[tough shale]

Add both of them up

You'll get the same answer

[violet dune]

i see

[tough shale]

That simplifies things, so moving to x^x²,

Its derivative would be?

[violet dune]

so definition of derivative of $x^x = x^x(\ln{x}+1)$

[grand pond]

Krish

[tough shale]

Yes

[violet dune]

then derivative of x^x^2

would be $x^{x^2}=x^{x^2}(\ln{x^2}+1)$?

[grand pond]

Krish

[tough shale]

No

[violet dune]

where did i go wrong

ohh wait

[violet dune]

$\frac{1}{y} * y' = 2x\ln{x} + \frac{x^2}{x}$?

[grand pond]

Krish

[violet dune]

therefore $\frac{dy}{dx} = x^{x^2}(2x\ln{x} + x)$

[tough shale]

Yes, that's right

[grand pond]

Krish

[violet dune]

okay im getting it now

[tough shale]

Or u can use that little trick there

[violet dune]

yeah i was just about to look at that, i didnt open the spoiler because i wanted to do it myself first

[tough shale]

If you're confused,
Treat the exponent like a constant, differentiate, treat the base like a constant, differentiate, and then add

[violet dune]

thats still a little bit confusing ill start using that once i get more comfortable with it, but for now ill try the long way

[tough shale]

Now we've already reached your required answer 😅

[violet dune]

now since the derivative of 1 is 0, we can say that $\frac{dy}{dx} (x^{x^2+1}) = x^{x^2+1}(2x\ln{x} + \frac{x^2 + 1}{x})$. is that correct?

[grand pond]

Krish

[violet dune]

theres your 2xlnx that you like so much lol

[tough shale]

I don't think u would need this

[tough shale]

This is ur question itself

[violet dune]

wdym?

[tough shale]

.

[violet dune]

thats 2xlnx + x though, not 2lnx + 1

[tough shale]

$\int x^{x^2}x(2\ln x+1)dx$

[grand pond]

Suika

[tough shale]

Take the x inside the brackets

$\int x^{x^2}(2x\ln x+x)dx$

[grand pond]

Suika

[violet dune]

oh

so we get $\int_1^2{x^{x^2+1}(2lnx+1)dx$

[grand pond]

Krish
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[violet dune]

so the integral is just $x^{x^2}$ from 1 to 2

[grand pond]

Krish

[tough shale]

Yes

[violet dune]

so 16-1=15

[tough shale]

Yes

That's correct

[violet dune]

ok that makes so much sense thank you

[tough shale]

No problem

[violet dune]

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[inland patio]

Let ${u,v}$ be an ordered basis in $\mathsf R^2$. Then it is called right-handed if $u$ can be rotated counterclockwise through an angle $\theta$ ($0<\theta <\pi$) to coincide with $v$. Define the orientation $\operatorname{O}$ of ${u,v}$ to be $$\operatorname{O}\begin{pmatrix}u\ v\end{pmatrix}=\frac{\det\begin{pmatrix}u\ v\end{pmatrix}}{\left |\det\begin{pmatrix}u\ v\end{pmatrix}\right|}.$$
I'm working the following exercise: ${u,v}$ is right-handed if and only if $\operatorname{O}\begin{pmatrix}u\ v\end{pmatrix}=1$. \
I think I've managed to prove the forward direction, using the fact that a rotation counterclockwise by angle $\theta$ is given $\mathsf T_\theta(a,b)=(a\cos\theta-b\sin\theta,a\sin\theta+b\cos\theta)$. I don't know how to proceed with the backward direction. If $u=(a,b)$ and $v=(c,d)$, then we have $ad-bc>0$ since $\operatorname{O}\begin{pmatrix}u\ v\end{pmatrix}=1$. How do I proceed?

[grand pond]

psie

[floral apex]

backwards direction assuming O(u,v)=1?

[inland patio]

indeed

[floral apex]

maybe you can do contra and cases? hrm

actually no cases

assume theta over or equal pi

[inland patio]

ok, so prove if {u,v} is not right-handed, then O(u,v) equals -1 (since it can only equal 1 or -1). But what does it mean for {u,v} to not be right-handed?

[floral apex]

clearly if u,v is a basis for R2 then neither is 0

so both have an angle in a meaningful sense

so it must be that angle(u) - a = angle(v) for some a

you can reformulate this into either (u,v) is righthanded or lefthanded

with ...is chiral the word? maybe mirrored definitions along the domain of the rotation angle

does that make sense?

[inland patio]

well, I don't understand what you mean by

you can reformulate this into either (u,v) is righthanded or lefthanded

[floral apex]

it must be u can be rotated SOME amount to overlap v

if that amount is between 0 and pi, its righthanded

if its between pi and 2pi its left

[inland patio]

ok

[floral apex]

so ... i think this is enough maybe

once you have the angle domain

and your T

[inland patio]

but sorry if I'm silly... where did you use that O(u,v)=1?

[floral apex]

you dont, you assume its not 1

[inland patio]

right, we are doing the contrapositive

[floral apex]

and show its ~(right handed)

[inland patio]

we need to show O(u,v)=-1 though, right?

[floral apex]

theres some edge case stuff i think

[floral apex]

i dont think so

but you could

just use abs defn right

the piecewise one i mean

[inland patio]

so you are not proving the contrapositive then? because the contrapositive of O(u,v)=1 implies right-handed is left-handed implies O(u,v)=-1, or?

[floral apex]

contrapositive is assume O(u,v)!=1 show (u,v) isnt right handed

[inland patio]

the contrapositive of A -> B is not B -> not A?

[floral apex]

if you do the edge case and some defns and stuff this is O(u,v) = -1 implies (u,v) is lefthanded

[inland patio]

and O(u,v) can only equal 1 or -1

ok

[floral apex]

oh wait

did i get mixed up

yes lol

youre right, assume left handed

sorry its been a day

[inland patio]

no worries

assume left-handed and prove O(u,v)!=1

[floral apex]

yea

which i think having the domain does?

working with the signs of the trig fns

[inland patio]

so {u,v} is left-handed if u can be rotated in an angle theta (pi < theta < 2pi) to coincide with v? My book only says that a right-handed coordinate system is when u can be rotated in an angle 0 < theta < pi to coincide with v, and otherwise it is left-handed. So it doesn't give a definition really.

But it makes sense if this is the definition of left-handed.

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[fringe cairn]

I have this matrix, how do I solve it for V1, V2, V3 from the matrix. Also please tell me generally how to do it, I need to show the calculations. The k’s are the magnitude kilo. If you can’t read something lmk.

[fresh sparrow]

by hand??

[fringe cairn]

🙂

[nova yoke]

well v1 = s, to start with

you can do row reduction to find the others

[fringe cairn]

I buit the matrix

[tawny apex]

so you multiply each row of the first matrix with the coloumn of the second matrix and then add it

[fringe cairn]

like $0 = V2 * ( -1/(2.2k) + 1/(2.2k) + 1/(3.9k) - 1/(3.9k))$?

[grand pond]

unrushed

[tawny apex]

uhh no

wait lemme show you

[tawny apex]

i did one of it

[fringe cairn]

i feel like this is a back step from my equations

[tawny apex]

uhh you have to solve for v1,v2 and v3 so youll have to do this

[tawny apex]

you want to solve these equations right?

so why make a matrix?

ohh i think i know what you mean

that process would be lengthy tho

[fringe cairn]

That is kinda how the prof taught it that I would not need to solve the matrix, but to just make the matrix. Now I need to solve it to solve for another piece of the puzzle

I'm open to any solutions

[tawny apex]

youll have to first find the inverse of the first matrix

so lets say the first matrix is A, the one with v1 is V and the last matrix is B

[tawny apex]

ohh

yeah then just solve the equation from here

it would be faster

[fringe cairn]

my first thought is replace all v1 with 5

[tawny apex]

so we know v1 is 5

yeah

[fringe cairn]

set each other equal to each other

[tawny apex]

then we'll have 2 equations with 2 variables and ten we can solve it

[tawny apex]

yeahh

but youll have 2 variables

so it wont do much good

cause we'll need 2 equations to solve

[fringe cairn]

my other thought is solve for V3 of the first equation and place it in the second for V3

i started to do that earlier, but got scared of the algebra

[tawny apex]

yeah

this is better

[tawny apex]

the fractions are a little scary yeah?

convert them first

[fringe cairn]

like common denominator?

[tawny apex]

yes

and also remove the decimals by multiplying the numerator and denominator with 100

the numbers are bad lol

[fringe cairn]

90.7335k as a common

then 90733.5

[tawny apex]

what's k?

[fringe cairn]

kilo

[tawny apex]

oh ok ok

is 2.24*3.94 equal to 90.7335k?? idts

8.8256

[tawny apex]

where'd you get this from?

ohhh you are doing the third equation

i suggest you do the second one

cause it would be easier

[fringe cairn]

8.58k

[tawny apex]

find v2 in terms of v3 and plug it in the third equation

[tawny apex]

no, its 8.8256

or maybe i calculated it wrong

[fringe cairn]

i think you think my k are 4's

[tawny apex]

ohh lol ok yeahh

yeah so now our equation becomes $3.9k(v2-5)+ 2.2k(v2-v3)=0$

[grand pond]

Aria

[fringe cairn]

over 8.58k

[tawny apex]

not needed

cause its equal to 0

[fringe cairn]

i see

[tawny apex]

and 8.58 is not equal to zero, thus the numerator has to be 0

now solve this, and derive v2 in terms of v3

you can also derive v3 in terms of v2 but it would make it difficult for us as the third equation has more no. of v3 variables

[fringe cairn]

$V3 = (3.9k(V2-5)+2.2k(V2))/(2.2k)$

[grand pond]

unrushed

[tawny apex]

derive it in terms of v2

itll be easier later

so

$v2=(19.5+2.2v3)/6.1$

[grand pond]

Aria

[tawny apex]

im not writing k for now

[fringe cairn]

yes

[tawny apex]

also we could multiply 10 in the num and denominator

giving us $v2=(195+22v3)/61$

[grand pond]

Aria

[fringe cairn]

ok

[tawny apex]

plug this in the third equation

the calculation is so baddd

lol yeah

[tawny apex]

yeah

what was the common denominator you found for this??

[fringe cairn]

90.7335

[tawny apex]

so

$19.305(v3-5)+23.265(v3-u)+27.495v3-604.89=0$

[grand pond]

Aria

[tawny apex]

just make sure my calculation is correct or not

and multiply by 1000 here

giving

$19305(v3-5)+23265(v3-u) +27495v3-604890=0$

here u is our value of v2

[fringe cairn]

yes

[tawny apex]

this is so bad ughh

who gives calculations like this:((

anyways put the value of u and solve that too

[fringe cairn]

do you understand the context?

[tawny apex]

uhh its a physics question no?

[fringe cairn]

basically

yes

[tawny apex]

i never have seen this much difficult calculations in physics and thus its making me wonder if we are correct

[fringe cairn]

its a circuits class

[tawny apex]

can you share the question??

you are applying KVL??

[grand pond]

Aria

[fringe cairn]

probs should have used super position

[tawny apex]

uhh i think yeah

do you have to find Vl?

[fringe cairn]

VL is the final bit of the puzzle

[tawny apex]

mhm

we should have just used the matrix atp

[lethal steeple]

Do you know the AX=B method for matrix?

[tawny apex]

i thought solving this would have been easier but now we'll have to use matrix

[lethal steeple]

Do you know how to find inverse of a matrix?

[fringe cairn]

no

[lethal steeple]

Oh

Do you know crammer's rule then?

[fringe cairn]

no

I've only used matrixes once

[tawny apex]

ohh

yeah then youll have to use algebra

[lethal steeple]

This might help.
https://youtu.be/UfwXTMygeVs?si=zR2c5WY0UNmB_682

[fringe cairn]

Ok thank you

[fringe cairn]

Ok thank you

[lethal steeple]

But the calculation is still gonna be lengthy

[tawny apex]

yeah it has some pretty bad calculations

[fringe cairn]

Ok

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[woeful turret]

can someone explain what they did?

[fallow scarab]

which line

[woeful turret]

after they expanded the determinant

[fallow scarab]

expanding along R_1 means they took the coefficients of the first row and multiplied them by the sub determinants

[woeful turret]

yeah no i got that

after that part

[woeful turret]

the factor, then the homogenous expression of degree 3

[woeful turret]

yes dude i know how to expand a determinant 😭

[fallow scarab]

then what's your question

[woeful turret]

after they expanded the determinant

and got 0

what did they do to prove that its equal to the rhs

[fallow scarab]

did you simplify this

[woeful turret]

its just 0 right

[fallow scarab]

yes

,w simplify -2a(-4ac - (c-a)^2) - 0 + (a+c) (0-2a(c+a))

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[fallow scarab]

does that answer your question

[woeful turret]

i dont think u understood

my question

this part

and just because the determinant = 0 how did they prove b+c and c+a are factors?

[fallow scarab]

$a+b$ is a ``factor" of $\Delta$ when $a+b = 0$, then $\Delta = 0$

[grand pond]

riemann

[fallow scarab]

just like how $x+1$ is a factor of $f(x) = (x+1)(x+2)$ because $x+1=0$ implies $f(x) = 0$

[grand pond]

riemann

[woeful turret]

oh so they did the samething for b+c and c+a and that also yielded determinant = 0 so thats how they are the factors?

,w det{{r-1,-1,-5},{-2,0,3},{-6,0,6}}

,w det{{r-1,r-2,r-6},{2r-4,2r-6,2r-11},{3r-9,3r-12,3r-18}}

[grand pond]

Results provided by WolframAlpha

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Requested by yoda_576

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[woeful badge]

hey

[woeful badge]

Back at it with this annoying question

Solved it but there has been a dispute. The highlighted part in yellow is where the problem is. My friends are saying it’s 11 - 4.9 and others including me are saying 10 - 4.9

Which is correct? Is it 10 or 11?

[last arch]

11

[woeful badge]

How sure are you?

[last arch]

Uh decently sure

What's the justification for 10

[woeful badge]

Negative 5 to positive 5 is ten

[last arch]

It says there -5.5

[woeful badge]

Woah

[last arch]

I mean am I missing something or

Even though the curve itself intersects x=5

The whole area includes the rectangle from -5.5 to 5.5

[woeful badge]

Yeah I said woah cause ive been seeing 5

maybe casue the intersection or wtv

Just wanna get confirmation that this derivative result i got is correct, no need to check anything else

just look at the dn/dt

[midnight plank]

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[lusty cave]

hey how do i solve this kind of problem

[placid spoke]

Area of a triangle is

$A = (1/2)ab\sin(C)$

[grand pond]

doaby

[lusty cave]

thx

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[green mesa]

the combination of linear and rotational movement is throwing me off, what's the first step?

[green mesa]

<@&286206848099549185>

[sharp coral]

you can conserve linear momentum to find his linear velocity and conserve angular momentum to find his angular velocity

[green mesa]

really?

I'll try it

It worked

thanks

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[regal barn]

without graph is there any method?

[rose lake]

i havent done this before, but i think intuitively the answer should be 2

because theres 3 parameters. its the points connecting each function that are not differentiable (unless the function happens to smoothly transition to the next one)

[thorny cove]

My prep-school math teacher always recommended using graphs for questions like these

https://www.desmos.com/calculator/semlogrbbb

[regal barn]

can i use one thing here is that if i make equal three of these functions

so i got three points

x=1,7/4,-9/4

[spare tiger]

there can be more than 2 transitions by the way, like consider max(x^2, 1)

[rose lake]

oooh i didnt think of that

one of the functions can be put between a single function.. thats kind of annoying

[spare tiger]

yup

[rose lake]

idk how you would solve that rigorously

[spare tiger]

the best way to do this I think is to just calculate when 2 of the 3 are equal

[regal barn]

please show

[thorny cove]

the easiest way would be to see where the graphs are intersecting

since that's where the function would get a sharp point

[thorny cove]

hence you get 2

x = -3 and x = 1

[regal barn]

yes i was saying same thing

[rose lake]

i was right but for the wrong reason lol

[thorny cove]

ah, I see

[thorny cove]

lmao

[rose lake]

one of the functions doesnt even play a role..

sad

[thorny cove]

If you wish to do it without making a graph, just equate the equations

9 - 4x = 2x^2 + 3

start with that

then verify with 9 - 4x = 4x + 1 and 2x^2 + 3 = 4x + 1

[rose lake]

what happens if one of the intersections doesnt play a role either though

[thorny cove]

wdym?

[rose lake]

something like this

[thorny cove]

well if one of the function is not intersecting, you can't really get the maximum properly iirc

I mean, y is a function of the maximum of those three equations

[rose lake]

if i put them in the max function it just spits out the parabola lol. i guess its obvious in this case but it might not be in other cases

[thorny cove]

yes, that'd just become y = x^2 + 10

[thorny cove]

lmao I just realized that I had this question in a test two years ago

so... do I just close this thread or...?

[rose lake]

whatever your heart desires

[thorny cove]

oh alr lol

.close

oh yeah it's upto the creator

damn

[midnight plank]

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[midnight plank]

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[earnest forge]

A cabinetmaker uses cherry wood to produce 5 desks each day. Each delivery of one container of wood is $5000, whereas the storage of that material is $10 per day per unit stored, where a unit is the amount of material needed by her to produce 1 desk. How much material should be ordered each time, and how often should the material be delivered, to minimize her average daily cost in the production cycle between deliveries?

[earnest forge]

I don't know where to begin

[vague remnant]

gotta create formulas, then find derivative, then minimum

[midnight plank]

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[earnest forge]

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[last slate]

howw

[last slate]

I tried doubling the lengths to get 2 diameters but then I am lost

The answer is A but how

[lyric charm]

call the radii of the circles r_A, r_B etc

then AB = r_A + r_B

BC = r_B + r_C

and so on

[midnight plank]

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[chilly cobalt]

"Consider a trapezoid with diagonals with length 5 and 3. If the length of the line connecting the bisectors of the parallel sides is 2, what is the area of the trapezoid?"

[chilly cobalt]

i have no idea what to do..

okay so

ive been stuck on this for a bit

i draw BE parallel to AB

and another line parallel to the line connecting the midpoints at B

[chilly cobalt]

and this line bisects DE as well

also ive noticed that BDE and ABCD have the same area

so all thats left is to find the area of BDE

how do i do that?

[chilly cobalt]

<@&286206848099549185> ?

[midnight plank]

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@chilly cobalt Has your question been resolved?

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[cedar mason]

[violet rivet]

🔺️FEB'
½ • 2 • 3 = ½(x + y)h

Trapezium area
= ½(2x + 2y)h = ½ • 2(x + y)h = (x + y)h

[carmine sigil]

What is the actual question?

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[cedar mason]

!nosols .

[midnight plank]

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

[carmine sigil]

@cedar mason that guy is OP

[violet rivet]

here

[carmine sigil]

Oooh

[violet rivet]

oh it's closed

[carmine sigil]

Nevermind

!nosols

[midnight plank]

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

[cedar mason]

xD

hilarious

[carmine sigil]

I'm a funny guy

Funny looking, funny smelling

[cedar mason]

.close

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[gusty flame]

how do i approach this problem

im guessing i should use the root test but how would that even work with the n^2

[gleaming latch]

what problem lol all you gave is a summation

are we proving it diverges?

[lyric charm]

can you write out what you need to work out for the root test, without worrying about how that would work

just apply the test as-written without thinking about how to simplify it

[gusty flame]

conv or div

[gusty flame]

well when i do the root test

[floral apex]

you could use the root test its a fools errand tho

why not check easier requirements firs

[gusty flame]

honestly not sure how to

[gusty flame]

whats a fools errand

[cedar mason]

waste of time

[floral apex]

like check nth term, or do a naive comparison first

[gleaming latch]

well it obviously diverges right

[gusty flame]

yes ofc

[floral apex]

its not obvious to them i dont think

oh

[gusty flame]

i mean i can see it divs

[floral apex]

well yea so

check the nth term

[gusty flame]

so (inf/inf+1)^infinity?

how do i evaluate that?

[floral apex]

we can take log right

[gusty flame]

not sure how to

[floral apex]

im almost certain lemme make sure

[gusty flame]

no problem

.close

[midnight plank]

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[floral apex]

huh

i guess it does go to 0

sorry 😭

[gusty flame]

no problem bro

[floral apex]

@gusty flame you still around?

.reopen

[midnight plank]

✅

[floral apex]

i hit it with the root test

if youre curious

[gusty flame]

sure what does it look like

[floral apex]

so you get $\lim \qty( \frac{n}{n+1} )^n$

[grand pond]

jan Niku

[gusty flame]

right

[floral apex]

this is $\lim \qty(1- \frac{1}{n+1} )^n$

[grand pond]

jan Niku

[floral apex]

i had to do some approximating from here

this is $\lim \qty(1- \frac{n}{n+1} + \frac{n(n-1)}{2(n+1)^2} - \dots )$

[grand pond]

jan Niku

[floral apex]

approximately, as long as 1/(n+1) is small

[gusty flame]

what a weird problem

[floral apex]

for large n, all of these falling products are just 1

if you believe that?

[gusty flame]

yes

[floral apex]

like $\frac{n(n-1)){(n+1)^2}$ is just 1

[grand pond]

jan Niku
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[floral apex]

annoying

mb u can read that

[signal hawk]

Pretty sure you can't do this when the exponent is variable

[floral apex]

so this is actually $\lim _{n \to \infty} \sum \frac{(-1)^m}{m!}$

[grand pond]

jan Niku

[floral apex]

ahhh yu might be right

i was thinking all this approximation is not needed

its a known limit

im sure of it

[floral apex]

because this is 1/e

but..its not important

[signal hawk]

Yea just take log(exp(...)) and youll get that I think

[floral apex]

you mean of this

?

[signal hawk]

yea

[floral apex]

@gusty flame

so, its actually convergent

wacky

[gusty flame]

yeah weird problem

[midnight plank]

@gusty flame Has your question been resolved?

[midnight plank]

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[sweet shuttle]

how can I isolate a variable here -3x+4y=2

[sweet shuttle]

I could add +3x but that would give me 4y=3x+2

then I could divide by 4

[fathom atlas]

Use rules of equations

[sweet shuttle]

but then I would have y=3x+2/4

[fathom atlas]

Parentheses...

[sweet shuttle]

y=(3x+2)/2 ?

[fathom atlas]

4..

[sweet shuttle]

y

?

[fathom atlas]

/4

[sweet shuttle]

this is part of a bigger question btw here is the full question I basically have to get this to y = something and use that to solve and get the factors

Give all the solutions for the nonlinear system.

^ that is the instructions

[sweet shuttle]

oh yeah mb

[wanton rock]

here a hint. it's supposed
4y = (3x + 2)
the parentheses is important

[sweet shuttle]

but then I would have y unc I would just have an equation

[wanton rock]

"unc l"?

[sweet shuttle]

unc=uncle

[dusty portal]

He’s calling you old

[wanton rock]

[dusty portal]

It’s the whole point anyways.

[wanton rock]

tbh i still don't understand

[dusty portal]

You can just subsitute into the first equation, no?

[gritty bane]

Yes

[sweet shuttle]

but where would I put 4y=(3x+2) if I dont have x or y ?

[wanton rock]

move the 4

[sweet shuttle]

y=(3x+2)/4

[dusty portal]

Well, isolate y (which you did already), then plug it into the other equation.

[wanton rock]

to the right side

[sweet shuttle]

sorry typo

[wanton rock]

congrats 🎉

[sweet shuttle]

unc I just fucking had that and u told me not to use it ?

[dusty portal]

!volunteers

[midnight plank]

Helpers are just people volunteering their time to help you. Be polite and patient.

[dusty portal]

Just be warned

[wanton rock]

tbh i don't really understand his point

[sweet shuttle]

I was being respectful just a question

[dusty portal]

I get it, it’s just rubs off as rude

[sweet shuttle]

why did u tell me to use 4y=3x+2

when I had y=3x+2/4

[dusty portal]

Anyways, yes, you had it right.

[wanton rock]

because this isn't same as
y = (3x + 2) / 4

[sweet shuttle]

ahh ok thanks

ngl now im confused again I have x^2+(3x+2)/4=5

do I multiply by denominator ?

[dusty portal]

(Second term is squared!)

[dusty portal]

Yeah.

It ends up being a bit nasty, but not too bad.

[sweet shuttle]

thanks @dusty portal and @wanton rock ur not unc status I take it back

.close

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[last slate]

Hi

[normal prawn]

What's ur question

[dusty portal]

!da2a

[midnight plank]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[last slate]

The Gauss Markov assumption of random sampling

Does it imply Cov(Yi,Yj)= 0?

Where Yi, Yj are dependent variable values from the data

[cedar mason]

nvm idk

[polar star]

,w Gauss Markov assumption of random sampling

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by _mia_34

[polar star]

could you state the assumptions

nevermind

[fallow scarab]

https://en.wikipedia.org/wiki/Gauss–Markov_theorem

[last slate]

Random sampling assumption states that we are working with a completely random population and only one population

If there's any clarification needed pls tell me

<@&286206848099549185>

[cedar mason]

try #probability-statistics

[last slate]

It feels like it should be true that if we have a random sample, then it's individual Y terms are uncorrelated

[loud heath]

Hlo

[last slate]

Reiterating clearly: The Gauss Markov assumption of random sampling, Does it imply Cov(Yi,Yj)= 0? Where Yi, Yj are dependent variable values from the data.

It feels like it should be true that if we have a random sample, then it's individual Y terms are uncorrelated.

But then that implies autocorrelation by itself. Which seems wrong that one GM assumption would imply the other.

[cedar mason]

!occupied

[midnight plank]

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).

@last slate Has your question been resolved?

[untold willow]

Huh

[last slate]

.close

[midnight plank]

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[molten bay]

Anybody check this solution

[midnight plank]

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[twin ridge]

Can someone explain this to me?

[twin ridge]

I've watched 6 videos on the subject and tried reading the textbook, but none of them have even come close to telling me wtf this beta thing is, let alone how to find it.

.close

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[dusty portal]

Test

Yep it’s fucking dead

<@&268886789983436800> Don’t know if this is the right ping but the bot is down right now

Sorry if I shouldn’t have pinged

Aa

[dusty portal]

There we are

Back online

.close

[midnight plank]

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[grim vector]

Ain't no way bro is meta tester

[dusty portal]

?!

Me and bonk noticed it

I was just seeing if it was really dead

[fathom atlas]

Bot needs rest too

[grim vector]

Nothing bad to this

[buoyant yoke]

you dont need to do this much testing to see that the bot is dead

just try a few commands and thats al you need lol

[dusty portal]

Nah I’m like that

Paranoid

[buoyant yoke]

and pinging mods is unnecessary

[midnight plank]

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[raven hawk]

Is my calculator broken? Here's what I'm looking at

[raven hawk]

But I keep getting 40.814.....

(actually, my problem uses 2.75 instead of 2.8, but still)

[fallow scarab]

,calc 1/2 * pi * 3^2

[grand pond]

Result:

14.137166941154

[fallow scarab]

,calc 4 * 3 / (3 * pi)

[grand pond]

Result:

1.2732395447352

[fallow scarab]

oh you didn't use parentheses in your calculator probably

and similarly for the last term with 3pi in the denom

[raven hawk]

I've tried with and without parentheses.. tried each operation individually instead of all on a single line..

[fallow scarab]

then you're inputting the parentheses incorrectly

[fallow scarab]

follow this

[raven hawk]

oh, I see what it's doing

thanks

.close

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[uneven sandal]

$I = \int_\mathbb{C} d^2z \, \omega(z) P_j(z) \overline{P_k(z)} z$

[grand pond]

tobi

[uneven sandal]

can I evaluate this for general orthogonal Polynomials with weight w?

because the z term is kinda annoying and I dont know how to deal with it

found some recursion formula on wikipedia

.close

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[exotic basalt]

angle a?

[midnight plank]

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[exotic basalt]

rn u cant do sine rule

wait so then u have b already?

or did u that using sine rule

nice

then use sine rule now

to find A

wait hold on

no use cosine rule again

both works

ah

ok

yeah

so its like

recall how a sin graph works

it goes up to 1 at 90 then goes back down to 0 at 180

correct?

so if ur value u had was like 75

on the other side of the sine graph, 105 would also be a valid value

smth like this

so it honestly depends if the question was asking u to find an acute or obtuse angle

but a better way to bypass that ambiguity is using cosine rule

since cosine goes from 1 to 0 at 90 then 0 to -1 at 180

no overlaps

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[exotic basalt]

yeah we'll continue here

would u agree that sin(75) and sin( 105) would output the same answer

like if y=sinx, it would be the same y value

so thats why our calculator is a bit confused

it gives the value closest to 0 instead of what we want

so imo i'd rather use cosine rule as it removes that confusion

always

cuz angles in a triangle can never get past 180

and from 0 to 180 the cos graph is never symmetrical

np

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[meager ore]

Let $p$ be a prime number, $k \in \mathbb{Z}^+$, and $G$ a cyclic group of order $p^k$. Prove that $|\text{Aut}(G)| = p^{k-1}(p-1)$.

[grand pond]

Halex

[meager ore]

I know $p^{k-1}(p-1) = \phi (p ^k)$ is the number of generators of $G$

[grand pond]

Halex

[meager ore]

And also, $f \in \text{Aut}(G)$ implies $\langle f(g) \rangle = G$ where $\langle g \rangle = G$

[grand pond]

Halex

[gaunt plume]

Do you know what the automorphisms of this group look like?

That's probably the best place to go next

[meager ore]

Every element in G is a power of g. The image of g under some automorphism of G is also a generator of G

Is it wrong?

[gaunt plume]

No that's right, that's just a rephrasing of this

You can be a bit more specific though

[meager ore]

How so?

[gaunt plume]

As in that's just a property that every automorphism has

What you want to try and do is find all of them

[meager ore]

Yup, I want to get a relation that allows me to show they are $p^{k-1}(p-1)$.

[grand pond]

Halex

[meager ore]

Could you give me a hint please?

[gaunt plume]

Sure

[gaunt plume]

See if you can prove the converse to this

or a converse, I should say

[meager ore]

it would not be surjective?

[gaunt plume]

Wdym

[meager ore]

for the converse, if we assume < f(g) > ≠ G, then | f (g) | ≠ p^k = | G |

[gaunt plume]

That's the contrapositive

Let f be some homomorphism

[meager ore]

Okay I'm dumb

[gaunt plume]

And let g be a generator for G

Assume that $\langle f(g)\rangle=G$

[grand pond]

depression

[gaunt plume]

Show that f is an automorphism

That's the converse

[gaunt plume]

Nw it's easily done

The contrapositive is just the same statement but written differently (assuming TND at least but let's not go there)

[meager ore]

if f an homomorphism by hypothesis?

[gaunt plume]

Yes, f is a homomorphism

Any luck?

[meager ore]

Yes

Took a while

Let $y \in G$. Since $\langle f(g) \rangle = G$ it follows $( f(g) )^k = y$ for some $k \in \mathbb{Z}^+$, which is the same as $ f(g^k) = y$ since $f$ is an homomorphism. Thus $f$ is surjective

[grand pond]

Halex

[meager ore]

Assume $f(x) = e$ where $e$ is the identity of $G$. We will show that $\text{Ker}(f) = {e}$. Since $x \in G$, we have that $x = g^m$ for some $m \in \mathbb{Z}^+$. $$f(g^m) = (f(g))^m,$$ so $m = |G| \cdot k$ for some $k$. Hence $$x = g^{|G| k} = (g^{|G|})^k = e^k = e.$$ Thus $\text{Ker}(f) = {e}$.

[gaunt plume]

G is a group, not a homomorphism, so ker(G) doesn't really mean anything

Well done though

[meager ore]

my bad

it should be fixed now

[gaunt plume]

You're still talking about ker(G)

G does not have a kernel

Your logic is good though

[grand pond]

Halex

[gaunt plume]

That's better

Good

[gaunt plume]

So now can you finish the proof of this?

[meager ore]

There's no need to prove that f is an automorphism then < f(g) > = G?

[gaunt plume]

Oh sorry I thought you were done with that proof

As in I thought you'd proven that already

Yes you need it

[meager ore]

well, this implication is a little bit harder than the previous one

how do I show < f (g) > = G?

[gaunt plume]

I'd do it by contradiction personally

But you don't have to, you can do it directly as well

[gaunt plume]

I thought you had it already because you said so here

[meager ore]

it was by intuition haha

[gaunt plume]

Lol nw

Your intuition is correct

[meager ore]

By way of contradiction, if $\langle f(g) \rangle \neq G$ this would mean that $f(G) \neq G$ which contradicts the fact that $f$ is surjective?

[grand pond]

Halex

[gaunt plume]

Ummm... yes but you haven't actually specified why this means that f(G) =/= G. You just said there's a contradiction and left it as an exercise for the reader haha

Sorry if that sounds mean but I'd probably give half marks

Really focus on the fact that g generates G

[meager ore]

I'm a bit confused now

[gaunt plume]

$\bigl\langle f(g)\bigr\rangle$ is the group generated by the element $f(g)$. $f(G):={f(g)\mid g\in G}$ is the setwise image of $G$ under $f$.

They are not the same

a priori

[grand pond]

depression

[gaunt plume]

So one not being equal to the whole group doesn't necessarily mean that the other isn't

If that makes sense

[meager ore]

Let $a \in G$, it follows $a = g^k$ for some $k$, then $f(a) = f(g^k) = (f(g))^k$.
since $a$ is arbitrary, $f(G) = { (f(g))^k : k \in \mathbb{Z}^+ } = \langle f(g) \rangle$

[grand pond]

Halex

[gaunt plume]

Good

wd

[meager ore]

since f(G) = G as we get our desired result

[meager ore]

Now, how does this help me prove this?

[gaunt plume]

Combine the two results you just proved

what does that give you?

[meager ore]

Not sure how to conclude

[gaunt plume]

result 1

[gaunt plume]

result 2

Combine them

[meager ore]

$f \in \text{Aut}(G)$ if and only if $ \langle f(g) \rangle = G$ with $ \langle g \rangle = G$?

[grand pond]

Halex

[meager ore]

Not sure how to use this

Could you help me how to conclude please? @gaunt plume

[gaunt plume]

Good

That tells you exactly what the automorphisms are and what they look like

You should be able to count them now

For simplicity fix a generator g

[meager ore]

I might be missing something, but I do not get the relation with the totient function

[gaunt plume]

It was one of the first things you said

[gaunt plume]

.

[meager ore]

Yep, but I didn't say I know why it is the order of |Aut(G)| 😔

[gaunt plume]

Ah okay

[gaunt plume]

That comes from this

There is one automorphism for every generator of G

[meager ore]

May I be missing another property of Aut(G)?

[gaunt plume]

That's what that statement means

Because homomorphisms are determined exactly by what they do to the generators of a group

Do you see why that's true?

[meager ore]

There are $\phi (p^k)$ generators of $G$

[grand pond]

Halex

[gaunt plume]

Yes there are

[meager ore]

And every automorphism is determined by the generators of G

Which are $\phi (p^k)$

[grand pond]

Halex

[meager ore]

Is this wrong?

[gaunt plume]

No that's right

Do you understand why it's true though?

[meager ore]

Okay, I see the point now

[meager ore]

What I don't get at all is how we can make the conclusion in a formal proof

[gaunt plume]

If $x = g_1^{a_1}g_2^{a_2}\cdots g_n^{a_n}$, then $f(x) = f(g_1)^{a_1}f(g_2)^{a_2}\cdots f(g_n)^{a_n}$ for any homomorphism f

[grand pond]

depression

[gaunt plume]

That's why

if g_1, ...., g_n is a generator set

You've got all the pieces already

You've just proven that the automorphisms are in bijection with the generators of G

[gaunt plume]

This is the bijection

[meager ore]

G has $\phi(p^k)$ generators and every generator $g \in G$ determines an automorphism. So $| \text{Aut}(G)$ has $\phi(p^k)$ elements

[grand pond]

Halex

[gaunt plume]

a unique automorphism*

But yes

[meager ore]

I get the point. But is it clear enough? Like how do I write it in a formal proof?

[gaunt plume]

That sounds formal to me

If two sets are in bijection, then by definition they are the same size

[meager ore]

I didn't prove it was unique did I?

[gaunt plume]

That's from this

A homomorphism is uniquely determined by what it does to generators

[lapis garnet]

Hi everyone

I need a help

[gaunt plume]

Yo

go to #help-47

[lapis garnet]

Ok

[meager ore]

I didn't know that 🤔

[gaunt plume]

Okay maybe I shouldn't've said it then haha

but for a cyclic group it's pretty clear

Because all elements are of the form g^k

so f(g^k) = f(g)^k

The only thing that matters is what happens to g

[meager ore]

I am a bit confused

[gaunt plume]

Do you know what a homomorphism is?

[meager ore]

Yes

[gaunt plume]

What is it?

[meager ore]

A function f:A->B satisfying f(ab) = f(a)f(b) where A and B are groups with some binary operations

[gaunt plume]

Exactly

It's a function that preserves multiplication

[meager ore]

I know the product <x> determines a subgroup

[gaunt plume]

Yes

But in this case, $G$ is a cyclic group, so {\sl every/} element of $G$ can be written as $g^k$ for some $k\in\mathbb N$

[grand pond]

depression

[meager ore]

What I don't get is the uniqueness

[gaunt plume]

Okay

Let's say $f$ and $h$ are two different homomorphisms such that $$f(g)=h(g)=g^k$$

[grand pond]

depression

[gaunt plume]

Since they are different, there is an $i$ such that $f(g^i)\ne h(g^i)$

[grand pond]

depression

[gaunt plume]

What is the contradiction?

[meager ore]

Weren't we talking about automorphisms?

[gaunt plume]

automorphisms are homomorphisms

but with extra conditions as well

f and h can be automorphisms if you want, it doesn't make a difference

This is a more general result

An automorphism is just an invertible homomorphism from G to itself

[meager ore]

Why does such index exist?

They are different and all of the elements of G had the form g^m

[gaunt plume]

if $f$ and $h$ are different, then there must be some element $x$ of the group such that $f(x)\ne h(x)$

[grand pond]

depression

[meager ore]

So there's at least one value where they differ?

[gaunt plume]

But $x=g^i$ for some $i$, because $G$ is cyclic

Yeah exactly

[grand pond]

depression

[meager ore]

f(g^i) = f(g)^i = h(g)^i

This is the contradiction?

[gaunt plume]

Yeah exactly

if f(g)=h(g) then f and h must be the same function

[meager ore]

What does this exactly provd?

[gaunt plume]

Uniqueness?

[gaunt plume]

.

[meager ore]

An endomorphism of a cyclic group is unique

[gaunt plume]

Well no not exactly

There are quite a lot of endomorphisms

But we're talking about endomorphisms such that f(g) = something in particular

[meager ore]

That something in particular are the homomorphism satisfying this

[gaunt plume]

No, that something in particular is any generator of G

[meager ore]

Could you tell me what the statement is again please?

I got lost

[gaunt plume]

Your question was whether there was a unique automorphism for every generator of G

And I just showed that the answer is yes

[meager ore]

If you suppose there are two different of them, why you assumed f(g) = h(g)?

[gaunt plume]

That's what we're showing isn't it?

We're trying to show that there's a bijection between automorphisms of G, and generators of g

[gaunt plume]

.

For any homorphism f, f is an automorphism if and only if f(g) is a generator of G

So for a fixed generator of G, there is an automorphism f such that f(g) = said fixed generator

And this shows it's unique

Look it's been 3 hours and I have stuff to do

But I have given you everything

Ping helpers if you need any more help

[meager ore]

Yep, thank you for your patience

[gaunt plume]

But read through this chat because everything you need is here

Good luck

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

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[copper ginkgo]

uh help?

[lyric charm]

how did you get 1000 as the number of terms

[lyric charm]

hmmm

[copper ginkgo]

or am i understanding the question wrongly

[lyric charm]

i dont think they would have expected you to do that

you were instead expected to use the alternating series error bound

ie that the error in a finite approximation of an alternating series is at most the magnitude of the first term thrown out

[copper ginkgo]

idgi.

never learn alternating series error bound before

[lyric charm]

wait what

was it not brought up in class

for a series $\sum_{n=1}^{\infty} (-1)^n b_n$, with partial sums $S_n$ and true sum $S$, you have $$|S_n - S| \leq b_{n+1}$$

[grand pond]

Ann

[midnight plank]

@copper ginkgo Has your question been resolved?