help-4

[open niche]

Question was how should i now replace k

In e^iπ.... Or in cos + i sin

Which is easier for simplifying into a real function like the expansion of sum

[worn sparrow]

its more intuitive to write e^{i pi} and use exponent laws imo

if you don't want to leave it as eta_k for some reason

[open niche]

Ok thanks

[worn sparrow]

actually

[open niche]

But like result is supposed to be real

[worn sparrow]

if you want to look for relationships to get rid of i appearing

you should consider conjugates

because

[open niche]

So somehow ak/-24 times ln(x-ak) + the conjacutive should be real

[worn sparrow]

z+z* is real for all compex z

[open niche]

Yes but thing is

The conjacutives have different factors

But im guessing they are just opposite

So should be good

[worn sparrow]

factors?

you mean coefficients?

[open niche]

Yes

like ak/-24 should be matching with ak overline /-24

[worn sparrow]

so put them back inside

[open niche]

Ok so i write it in e^iπ form

Can i somehow match Pairs before expanding sum?

Like sum 0 to 11 ak/-24 ln(x-ak) + ak* /-24 ln(x-ak*) prob

[worn sparrow]

you have to find the right choice of indexing

,w plot z^24=-1

[rocky lotus]

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Requested by gfauxpas

[worn sparrow]

no the roots, ugh

one sec

you can do it taebek

simplify this beast!

[open niche]

This correct ?

[worn sparrow]

oh i thought you wanted to try combining them before integration

[open niche]

Oh no

Thats much more effor

Trig sub needed

I just use complex ln

But i think its better with csc with eulers trig identity

Instead of e^i..

That way i have both directions

@worn sparrow this correct ?

[worn sparrow]

yeah but this can probably be simplified too

[open niche]

How

Should i type it as cos and sin

Before expanding

[worn sparrow]

no

[open niche]

Inside the integral

Yes?

It is needed

[worn sparrow]

$z+z*=2\Re{z}$

[rocky lotus]

gfauxpas

[open niche]

Yes

That

But integrate first fs

[worn sparrow]

my cat is trying to get my attention by blocking my pc one moment

[open niche]

Crazy

[worn sparrow]

no because

[open niche]

If you dont integrate you'll get integrals of the type Ax+B/Cx^2+Dx+E

Needs square completing and trig sub

[worn sparrow]

sorry i cant focus on my pc lol

GL

[open niche]

Ill just do it and send

For someone to review but i think i figured it out

[worn sparrow]

you can still write something lik

[open niche]

I wanna show it

Ill just replace trig form and cancel out

[worn sparrow]

btw you don't want to write "ln" here. Or at least I wouldnt

there's more than one convention, but

[open niche]

Wdym

ln

Complex ln

[worn sparrow]

you don't mean the principal branch of the real natural log

you mean any branch

but it' s not a big deal because it's all collected into the +C anyway

[open niche]

ln(x-ak) = ln|x-ak| + iarg(x-ak)

[worn sparrow]

I wouldnt write that, I would wirte
ln(x-a_k) = log|x-a_k| + i arg(x-a_k) because otherwise ln means something different on the RHS and LHS

but then again there are different conventions about this

[open niche]

Log though you dont mention base

[worn sparrow]

it's implied

[open niche]

Whats the diff

[worn sparrow]

there's only one log worth mentionin as soon as you use complex numbers

[worn sparrow]

because otherwise

ln means something different on the LHS and the RHS as I said

[open niche]

What is it

The difference

Idk what each means

[worn sparrow]

on the LHS it's a function of complex numbers, on the RHS it's a function of real numbers

[open niche]

Yes

And they are equal

Im using the complex ln

[worn sparrow]

"the" which one

[open niche]

And it will become real one after i match pairs

Since im part will go to 0

[worn sparrow]

it doesn't matter at the end of the day because it's implied by the +C but

there's nothing stopping you from choosing a different log for each factor

so nothing cancels out

alright nm, i 'll stop annoying you about it

[open niche]

Whats conjacutive

In e form

e^iθ* = e^-iθ?

[worn sparrow]

you tell me

how do you tell

?

[open niche]

Wdym

[worn sparrow]

i'm asking you to figure out the answer on your own

the conjugate of a number a+bi is the number a-bi, for real a,b

[open niche]

Well id just consider the function with cosx -i sinx equal to e^-ix

And proce its true

[worn sparrow]

bingo

correct

[open niche]

Ok so it is this

[worn sparrow]

i didnt want to hand the answer to you, i wanted you to figure it out

yes

for |z|=1

[open niche]

Im confused

I cant tell what cancels

,wr90°

[open niche]

Whats the command?

[worn sparrow]

,rot

,row ccw

.rot ccw

,rot ccw

idk

[open niche]

,rccw

[open niche]

You see anything i can do

I can factor out cos and have multiplication of ln but cant really do none with that

[worn sparrow]

what's that symbol inside cos and sin

pi?

[open niche]

Yes

[worn sparrow]

honmestly it's hard for me to think about that many terms at once

but trig identities about supplementary and complementary angles might be relevant

[open niche]

Great idea

I gtg do some ill check trig swaps later

[vale dock]

@open niche Has your question been resolved?

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[wind ether]

can i take the 2^2x as 4^x ?

[drifting hornet]

sure

[glass kelp]

that looks horrendous

[drifting hornet]

are you trying to simplify it?

[wind ether]

yeaa

i thought maybe i can do some log

[drifting hornet]

sin cos^-1 can be simplified

[glass kelp]

consider a triangle 👀

[drifting hornet]

I'd prolly start with that

[wind ether]

🤔

okay

pythagoras

?

[autumn whale]

draw out a triangle

[glass kelp]

yup

[autumn whale]

and think about what the cosine function is in terms of the triangle

[wind ether]

huhh?

HAan

got it

[autumn whale]

[wind ether]

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[winter crater]

This problem is giving me trouble. It asks specifically for an induction proof; but when going about it, I encounter a term that I can't get rid of

[wraith heart]

do you recognize that this term is positive

[winter crater]

Yes

[wraith heart]

apply that to your last line and this goal should be immediate

[winter crater]

I'm not following. How does that help?

[wraith heart]

you see this right?

[wraith heart]

do you see how it differs from the right side from the goal?

[winter crater]

Yes, it differs by that ka^(k-1)h² term

I also just noticed I'm missing an exponent of k for the kah term

[wraith heart]

do you see why if x >= y + z and z > 0, then x >= y?

[winter crater]

No

[wraith heart]

try adding y to both sides of z > 0

[winter crater]

Okay, I've got y + z > y then, where does the equals come from?

[wraith heart]

combine what you got with x >= y + z

[winter crater]

Right, that gave me x >= y + z > y

[wraith heart]

and u > v implies u >= v

[winter crater]

Right, because of inclusive or

Forgot about that

Thank you. I found where the missing exponent went and it worked out!

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[silk peak]

can yyou guys help me make a word problem

[autumn whale]

for?

[autumn whale]

so you just need to make a word problem that can be modeled and solved using a quadratic equation

projectile motion problems usually involve quadratics

[exotic token]

i got an idea but i think thats a cool project for you to think about!

[stark wedge]

can we even give any hints that would not amount to doing the problem for op

[silk peak]

no im just bad at logic based things 😢 like what would be an activity or somethign that would result in a quadratic

uh

visualization

[autumn whale]

projectile motion

[silk peak]

i mean

[exotic token]

since its a creativity based project i think that it can be pretty satisfying if you come with an idea

[autumn whale]

you can do something related to focii

theres a lot you can come up with

[silk peak]

whats focii

[autumn whale]

the plural of focus

have you gone over focus and directrix yet

[silk peak]

no

[autumn whale]

oh nvm then

[vale dock]

@silk peak Has your question been resolved?

[wild linden]

@silk peak one class of easy real world problems involving quadratics are anything involving tracking the position of accelerating objects.

[silk peak]

how

[vale dock]

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[tropic sun]

Hello! I'm taking my TEAS exam, and I need help with visualization on certain geometric shapes. In this problem, I already know the answer, but I guessed instead of sketching it out, and I REALLY want to be able to visualize it properly. Is it possible for someone to sketch a quick diagram of the problem and what they mean by "the perpendicular distance between the pair of opposite sides"?

[plush anvil]

I think

[lean chasm]

well wudnt the perpendicular from left to right be 45?

i originslly drew that out too, but dont see how L>R wud be 40.5

[cloud epoch]

its not perpendicular so no

the 30cm and 45cm sides arent perpendicular to one another

[lean chasm]

right

mb

[tropic sun]

My random ass guess was that I should divide the area by the other number. I don't even have a good explanation as to why, it was a random dice roll.

[lean chasm]

cant read

[upbeat crow]

oh is this the ultimate server for overlyintellectuals

[lean chasm]

i see what theydve done tho

dividing the area by 30

[tropic sun]

I think they divided the paralellogram into 4 rhombuses

I think

[cloud epoch]

Its just that the formula for area of parallelogram is base x height

so you can take any of the two bases/heights

[lean chasm]

i have no idea how to draw this tho

[cloud epoch]

and therefore you should have that 30cm x height = area

therefore the height is area/30

:)

[noble anvil]

Welcome

[tropic sun]

So maybe I don't understand parallelograms as well, because I'm a little confused as to how 30cm counts as a base...

[upbeat crow]

thank you!

pleased to be here

[cloud epoch]

any side can be the base

[tropic sun]

Either the diagonal or the length?

That is nutty

[cloud epoch]

all sides can count as a base, and then the height is the distance to the other side

[tropic sun]

ohhh

Wait so

[cloud epoch]

get it?

[tropic sun]

Let me see if I can sketch this

[cloud epoch]

i can on paint if you want

[lean chasm]

holdon

[cloud epoch]

might not be pretty

[lean chasm]

if u see it like this

[cloud epoch]

either multiply the two reds, or the two greens to get the area either way

[tropic sun]

yeah thats what I thought

[cloud epoch]

👍

[tropic sun]

They wanted that green length

[cloud epoch]

yup

[tropic sun]

That is so vaguely worded though. Ugh

I hate the TEAS exam.

Thanks for the help everyone!

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[spring jackal]

<@&268886789983436800>

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[small niche]

I'm planning on returning to uni this fall to finish my engineering degree which is basically just all 4 years anyway... I have been going from very early math up to uni and still have a while to go. However, I haven't been getting enough just algebraic manipulation practice and I know how important that is. Are there any compilations that just lay out all of the manipulation rules/properties in an easy to understand way ? If so that would be great

[small niche]

a lot of the implied things like a negative exponent is just a reciprocal so in exponential decay when your common ratio is 1/2 and you go backwards with a negative exponent I understand how that makes sense in that you're now dividing by 1/2. There are a lot of little things like that I forget though, does anyone know what I mean?

[fleet junco]

khan academy maybe

[small niche]

I appreciate the help

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[long shuttle]

not sure if i did the calculations here correctly

[long shuttle]

p(x) when x = 0 + p(x) when x = 1 right?

so 0.135 + 0.271?

[west cloud]

mhm, that seems right, but make sure you phrase that correctly: it's p(x) when x=0 and x=1

[west cloud]

[long shuttle]

ohh ok ok thank you!!

[west cloud]

!done

[vale dock]

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[long shuttle]

one more question I just want to checl

sorry

[west cloud]

mhm?

no need to apologize

[long shuttle]

it is a different question for the homework, do i have to make a new channel?

[west cloud]

no, it's fine

just post it here

[long shuttle]

this question

[west cloud]

-# I'm not sure about this one sorry

Wait a while, other helpers will come. If they haven't for 15 mins, you can ping @Helpers.

[long shuttle]

I went with this question because

0.3 - 0.5 = 0.20

then the standard dev is 0.0707 so

i divided 0.20 / 0.0707 which got me 2.83

[long shuttle]

oh sorry I hadn't sent the full question yet

[velvet lake]

!original

[vale dock]

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.

[velvet lake]

that's not the original. The original has stuff before

we need that context

[long shuttle]

thats the original

[west cloud]

might be because this question has multiple parts

we can see that by the phrase "larger bag of skittles"

[long shuttle]

this screenshot is the entire question sorry

[velvet lake]

this is a part of a bigger problem, and it's not the first part. We need the parts before

[long shuttle]

by that I meant the question with the options

[long shuttle]

there isnt any other parts

[vital pecan]

From the answers provided, it is sensible to assume there are 5 colors, and the color picked for a specific skittle fits a discrete uniform distribution

[velvet lake]

what comes immediately before of what you posted?

[long shuttle]

this question

this one i already got it tho

[west cloud]

are there any questions before this one that mention skittles?

[long shuttle]

nope

[west cloud]

hmm

[long shuttle]

we are just supposed to make an assumption

[velvet lake]

apparently skittles bags have 5 flavours

i guess they just force you to assume what the fuck is in a skittles bag

[vital pecan]

It is implicitly stated in the possible answers that we assume the expected amount of green skittles is 20% of the total. So we have that p = 0.20.

[velvet lake]

and you also have explicitly that you have 100 skittles in the bag

so n=100

so... run the simulator with that? i guess?

[long shuttle]

yeah i did

[velvet lake]

that's... not the value of p that you're supposed to use

[long shuttle]

oh my bad

[vital pecan]

Then remember to use the provided value for p̂

[long shuttle]

got it!! ty!!

omg

.close

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[south sorrel]

Okay, so I know what to do once I have the magnetic field vector

But I don't know how to get the magnetic field vector...

the equation I know just gives me the magnitude, but I need the direction as well

This is the equation I have, but it gives a scalar quantity

[neat locust]

The direction can be found based on whether your currents are parallel or antiparellel

[south sorrel]

How so?

[neat locust]

oh wait. Sorry I'm thinking force

[south sorrel]

😭

[neat locust]

For direction of B, use right-hand rule

[south sorrel]

outta da page, then?

I'm lost on what happens, though
Because I know that

$\int\frac{Lsin90}{r}$

[rocky lotus]

Blake

[south sorrel]

How does that become what that equation simpliefies to?

[neat locust]

I think it's into the page, actually

[south sorrel]

How do I know the direction of the L vector?

[neat locust]

It's the same direction of your current flow

[south sorrel]

Mm
That seems arbitrary

But intuitive

It's 9:30, let's go with it :)

@neat locust What's the angle between B and the vertical and horizontal wires in the rectangle?

I'm pretty sure the vertical are 90

I'm confused about the horizontal, though

[neat locust]

parallel would be 0, and antiparallel would be 180

[south sorrel]

So B is into the page

and the horizontal run in the x axis

I don't see how they're either parallel or antiparallel lol

Oh, wait

[neat locust]

B for I_1, yeah that's what I'm getting

[south sorrel]

Would they be 90?

Surely they're not both 90, though...

Well, why couldn't they, I guess

[neat locust]

90 with what

[south sorrel]

between the horizontal wires in the rectangle and the magnetic field

[neat locust]

Oh. Yes.

[south sorrel]

all the wires are 90 with B?

And if you don't mind, could you tell me if it's accurate to say that the magnetic field is greater at point B because the two wires add together for point B, but subtract (because the magnetic fields are in opposite directions) for point A?

[neat locust]

Is this an entirely different question?

[south sorrel]

:D

Well, I did say "if you don't mind"

feel free to abandon me

and make be completely fail summer school, and live under a bridge because I couldn't get my degree

[neat locust]

btw yes this is a loose way of saying the correct answer

[south sorrel]

I tried to do the right hand rule, but it seemed like for both points the wires contributed a magnetic field in the same direction

[neat locust]

You may want to mention which direction the magnetic field is going

[south sorrel]

At point B, the magnetic field is out of the page, due to wire 1

[neat locust]

Break it down case-by-case

I1 on A
I1 on B
I2 on A
I2 on B

Which directions are the magentic fields going in each case?

[south sorrel]

which way should the r vector be?

Towards the wire or towards the point?

[neat locust]

wire-to-point

[south sorrel]

I1 on A is into the page?

Now I2 on B

I'm thrown off, it seems like it's out of the page

Which lines up...

[neat locust]

You're saying I2 on B is out of page?

[south sorrel]

So I1 on A, then is into the page

[south sorrel]

that's what I thought...

[neat locust]

You are correct

[south sorrel]

okay

So the magnetic fields on B add

I1 on A is into the page

and I2 is out

so they subtract

okay

Then do the math for part b, then use some equation for torque for the explanation in part c

Thank you @neat locust
Have a good night my friend

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[south sorrel]

@neat locust I have another question :D
So, what distance, r, do I use for the vertical lines? It changes as you go along the line
(please don't say integration)

[neat locust]

this is why you need a line integral 🙂

[south sorrel]

😭

We don't have equations with any integrals for force, only for magnetic fields

F = IL X B

[neat locust]

Force would be the same thing

[south sorrel]

$F = I B sin \theta \int dL$

[rocky lotus]

Blake

[south sorrel]

Please tell me that's right, that would be so easy

[neat locust]

$\vec{F}=\int I\dd\vec{L}\times\vec{B}$

[rocky lotus]

SWR

[south sorrel]

I is constant, take that out

dL cross B is dL times B times sin theta

oh wait, is B not constant?

[neat locust]

it would depend

[south sorrel]

on distance

[neat locust]

B and L are functions of distance, yes

[south sorrel]

So what happens to our integral?

[neat locust]

It depends on the exact problem

[south sorrel]

the distance from the long wire to the top of the rectangle is 7cm

So the distance for B is dL + 7cm?

[south sorrel]

$\vec{F}=\frac{uI}{2\pi}\int I\dd\vec{L}\times\frac{1}{R}$

[rocky lotus]

Blake

[south sorrel]

$\vec{F}=\frac{uI^2}{2\pi}\int \dd\vec{L}\times\frac{1}{\dd\vec{L}+.07}$

[rocky lotus]

Blake

[south sorrel]

.reopen

[vale dock]

✅

[neat locust]

what the heck is happening here

[south sorrel]

I don't know, I tried...

I'm not good at integrating in physics

Calculus belongs in calculus 😭

Not in Calc based physics

[neat locust]

calculus was invented for physics

[south sorrel]

And this course is called "Calculus-based physics"

[south sorrel]

life was simpler when we lived in huts

who cared about velocity? Farm, raise children, die, day of judgment, resurrection

[neat locust]

Okay. First, what question are you trying to answer? You;ve been asking a few and I don't know where you are exactly

[south sorrel]

So, I'm back on the first one

I thought I was cooking with gas

then I realized I didn't know what distance to use for the vertical wires

Horizontal are easy, I've got R, I've got L and I, no problem

the R and L varies for the vertical, though

[vale dock]

@south sorrel Has your question been resolved?

[south sorrel]

.close

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[ashen viper]

help

[ashen viper]

I need someone to walk over this quesiton with me, its simple but i dont get it:

[stark wedge]

have you made any progress so far?

you're obv gonna need some kirchoff law magic

[ashen viper]

nah

i dont even where to start

any advice?

[craggy mist]

You won't need kirchoff

Calculate the equivalent resistance

I think, you already know how to do that, by looking at the ckt. From the back, or as they say.

[karmic yacht]

Are you familiar with series and parallel connections

resistors*

[ashen viper]

I'm not too sure, i went from coulombs law to this in a weel and have an my first exam tmr(today) i'm more lost than u think

cooked ggs

[ashen viper]

Yes

[karmic yacht]

do you know how to calculate equivalnt resistence

[ashen viper]

ye

but not too sure for htis

[karmic yacht]

alr look at 4r and 2r

[ashen viper]

for series u add them for parallel u do ^-1

[karmic yacht]

arent they in series

[ashen viper]

yea

how do u add them to make them a loop?

[karmic yacht]

so can you replace them with one single resister?

[ashen viper]

oh wait, so u add them, and than do parallel with the 3R right? and then after solve that by adding the regular R?

[karmic yacht]

yes

[ashen viper]

omg

[karmic yacht]

thats just for total equivalent resistance by the way

[ashen viper]

I got 30 Olm's but im not too sure

.close

[vale dock]

Channel closed

Closed by @ashen viper

Use .reopen if this was a mistake.

[karmic yacht]

0.2 A is not total current

[ashen viper]

.reopen

[vale dock]

✅

[ashen viper]

what does this imply?

[karmic yacht]

Did you check your answer with the key

[ashen viper]

there's no key

Couldnt* find it

[karmic yacht]

kirchoff rules are required

[ashen viper]

how do I apply that?

[karmic yacht]

Have you learnt the,

them*

[ashen viper]

the junction tule and the loop rule?

[karmic yacht]

I think youll need only the current sum rule

IN current sum = OUT current sum

into a junction

[ashen viper]

yea I remember, but how will that be applied here? after summing the resistance?

[karmic yacht]

At the junction between R , 6R and 3R the current splits

In going current is total current

Current going through 3R is given

[ashen viper]

imma be honest

I have no clue

[karmic yacht]

Draw the circuit keeping everything the same except the 2 resistors in series combine them to 6R

[ashen viper]

yea

[karmic yacht]

alr now theres a junction between the resistors labelled R, 3R and 6R right

[ashen viper]

so the original current will be 18V/R hat split to I for 3r and 6r right?

but then how do u solve for current three going to 6R?

[karmic yacht]

The resistance of 6R s double that of 3R right

So half the current flowing through 3R will flow through 6R

[ashen viper]

ooooooooh

so the total i is 0.6

[karmic yacht]

huh

how

[ashen viper]

nvm

i got it reverse

[karmic yacht]

do you want a different approach

[ashen viper]

lowkey

[karmic yacht]

You knw abput voltage law?

[ashen viper]

not too sure

[karmic yacht]

or do you wanna continue with current approach

[ashen viper]

IG stick with the current one

[karmic yacht]

Alr so current passing through 3R is 0.2 right

[ashen viper]

so the other half is 0.2 as well going 6R?

[karmic yacht]

No

[ashen viper]

oh ok.....

[karmic yacht]

The same current doesnt go to both

Because it is "tougher" to go through 6R than 3R

How much more "tough" is it to go through 6R compared to 3R?

[ashen viper]

twice

is it cuz V dont chnage throughout so its dependent on I

[karmic yacht]

So current through 6R will be half of current flowing through 3R right

[ashen viper]

so its 0.1

[karmic yacht]

so total current?

[ashen viper]

so 0.3

[karmic yacht]

alr now use equivalent resistance that you found earlier

V=IR

[ashen viper]

so R will be 60, 18/0.3

[karmic yacht]

are you sure

equivalent resistance is 3R

of the entire circuit

[ashen viper]

ooooh right

so its 20

dam

[karmic yacht]

You can verify this using voltage rule

[ashen viper]

so just voltage of each right and add them up?

[karmic yacht]

yes but choose the loop properly

You can use the voltage rule alone to get to the solution also

[ashen viper]

dang

yea I did get 18V by adding R and 2R after parallel

that made akot more sense, thanks for the enlightment, gotta go before i get no sleep

preciated

.close

[vale dock]

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[wispy verge]

How do you solve xlnx = 2i(pi)

[lyric bloom]

aln(b) = ln(b^a)

ln(x^x) = 2ipi

x^x = e^(2i*pi)

Try from here ig

Idk about solutions in C tho

Ig some branch of lambert W

Idk

[viscid spade]

e^(2i*pi) is 1

[lyric bloom]

Yes

[wispy verge]

that was the original problem

i put the ln of both sides

because idk i thought it would be easier

[wispy verge]

yeah

it is

[lyric bloom]

Ah so you are going in the wrong direction

[wispy verge]

but idk how to use lambert w

[wispy verge]

wait so like

what do i do then

[viscid spade]

Let u = ln(x)

[lyric bloom]

I mean yes you are going right but need one correction

[viscid spade]

To make it appear

[lyric bloom]

x^x = 1

That was the og question right?

Seems like a question from a bprp video 😭

[drifting hornet]

except x = 1 doesnt actually work

unless we allow multivalued log stuff

[lyric bloom]

I mean the original question is x^x = 1

I think

[drifting hornet]

oh

[lyric bloom]

He didnt include the n multiple after taking log

[drifting hornet]

so we are solving x^x = 1 in C?

[lyric bloom]

I think so

Otherwise it would be trivial ig

x^x = 1

Take log

xlog(x) = 2n(pi)*i

And use lambert W ig

[lyric bloom]

Its the inverse function of f(u) = ue^u

[wispy verge]

yeah i made it up like an hour ago 😭

[wispy verge]

I wanted to find complex solutions to this

not real ones

its quite obvious what the real solution is

[wispy verge]

do we need to

[lyric bloom]

Yes

[wispy verge]

I wanted to just assume n = 1

[lyric bloom]

Then you'll find a very specific solution

[wispy verge]

like i wanted to find the smallest complex solution too

[wispy verge]

exactly

i wanted to find the solution in the form x + yi

[lyric bloom]

There is no ordering in complex numbers

There is no "smallest" complex number

[wispy verge]

what about the modulus of the complex number

can't that technically be considered the "size"

[lyric bloom]

I'm not sure if n=1 will yield the smallest magnitude

[wispy verge]

ah

what will then

well could we even know

[lyric bloom]

Not sure how to check tbh but my guess is as n goes to infinity

[lyric bloom]

The value will be smallest

[wispy verge]

oh

nvm then

ig i just want

a specific solution

[lyric bloom]

Ohk

[wispy verge]

so like how do i use lambert w again

[lyric bloom]

Ig using different branches of lambert W, there'll still be infinite solutions to your problem

[wispy verge]

yeah ofc

but lets just say i set n = 1

then it wouldnt be infinite

[lyric bloom]

It will still be

Wk(2pii)

For diff value of k, there will be different solutions

[drifting hornet]

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

if you just want the numerical sols, this works

[lyric bloom]

It gives complex values too?

[drifting hornet]

blue is Im((x+yi)^(x+yi)) = 0 and red it Re((x+yi)^(x+yi)) = 1

[wispy verge]

yeah it gives everything

bruh this pattern is cool

but im guessing the biggest value that gives x^x = 1 goes to infinity?

[drifting hornet]

yeah, seems like it

[wispy verge]

hmmm

[drifting hornet]

goes to infinity in magnitude

[wispy verge]

i wonder what im gonna say as a solution to this tbh

i feel picking any complex solution would feel super random

[lyric bloom]

Well you made the question yourself

[wispy verge]

yeah true

idk i was just curious

[drifting hornet]

Pick 0

[wispy verge]

because i had a gut feeling x^x = 1 had non real solutions

.solved

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[faint ravine]

basically

you have these equations and u need to find the minimal value of 100x+10y+z

my question is

it says in the solution in the picture:"from the asumption 96=xy+yz+zx>=3x^2 where did we get this from?

<@&286206848099549185>

o yeah and x,y,z are natural numbers

[jovial edge]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[vale dock]

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[jovial edge]

And also, can you translate the problem

And solution

[faint ravine]

given the equations and x,y,z are natural numbers,find the minimum value of 100x+10y+z

solution:we can see that the equations are simetrical,without the loss of generality we can assume x<=y<=z

from the first equation ....=1680 <=>(96+x^2)(y+z)=1680
from the asumption we have 96=xy+yz+zx>=3x^2

[jovial edge]

Oh yeah, since we suppose $x \leq y \leq z$

Oops

[rocky lotus]

Alexis_Fx

[faint ravine]

my question is where do we hey the last inequality from(the one with the 96(on the bottom))

[jovial edge]

When $x=y=z$ we have $xy+yz+zx = 3x^2$

[rocky lotus]

Alexis_Fx

[faint ravine]

ok

thanks

.close

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[lime bone]

I’m so mad

[lime bone]

I’m so mad

A (1,0,0)

B(0,2,0)

Now project it into xy plane and tell me the equation

People

Find the normal first

It is (2,1)

No problem

[noble anvil]

the normal to what

[lime bone]

So the line can be written as 2x+y=r

r is 2

Is it how it is done

Please Mickey lord

Tell me

Any shortcut

Please

Mickey lord

Should I say Mickey lord or lord Mickey

[white marlin]

!original

[vale dock]

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.

[lime bone]

.close

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[noble anvil]

[jovial edge]

what was that

[lyric bloom]

That pmo

[gloomy harness]

what

happened

[white marlin]

lol

[gloomy harness]

bro went on a schizo rant

[noble anvil]

Fr

[red tulip]

https://tenor.com/view/better-call-saul-chuck-mcgill-chicanery-breakdown-gif-25422432

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[devout holly]

How do I know whether my LU factorisation is correct?

[devout holly]

There are really many answers depending on how I go about my Gaussian elimination... correct?

How do I check if it's correct? The answer key has a different L

[wild linden]

LU = A, you can check your work simply by doing the multiplication.

,w {{1, 0, 0}, {2, 1, 0}, {2, -2, 1}} . {{1, 1, 2, 4}, {0, -2, -3, -7}, {0, 0, -6, -11}}

[rocky lotus]

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[wild linden]

Looks like you made an arithmetic mistake at some point, the last row is incorrect.

[devout holly]

It's can't be multiplied tho....3x4 with 3x3

[wild linden]

3x3 with 3x4 can be

[safe fulcrum]

AB is not equal to BA for matrices

[devout holly]

The answer

[devout holly]

Yup

L is 3x3, U is 3x4

[wild linden]

In this case LU exists and is a 3×4 matrix. UL does not exist because of the incompatibility of the dimensions

[devout holly]

Ahhh my bad I saw wrongly

[devout holly]

Mhmm

[wild linden]

,w {{1, 0, 0},{2,1,0},{0,0,1}}.{{1,0,0},{0,1,0},{2,0,1}}

[rocky lotus]

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[wild linden]

,w {{1,0,0},{2,1,0},{2,0,1}}.{{1,0,0},{0,1,0},{0,-2,1}}

[rocky lotus]

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[devout holly]

I see

I'll double check!

[wild linden]

The error would need to be in the last row of the L matrix

And you performed the matrix multiplication correctly, which implies that perhaps the issue is with turning your row operations into matrix values.

[devout holly]

Is there any way to find L without multiplying 3 matrices?

I feel that it's ripe for arithmetic errors

I found the error, but the answer is still incorrect

Hmmm

[wild linden]

If you can find a way to make an upper triangle matrix with fewer row operations then you need to do fewer matrix multiplications.

[devout holly]

Wait is it definitely an error with L?

I know the middle 1 should be -1 instead

But the output is stil wrong

[wild linden]

Well, that's the most likely cause of just the last row of the product being incorrect

[wild linden]

Here

[devout holly]

I've checked multiple times but I can't spot the error

O.o

Strange

Ah i multiplied the matrices in the wrong order

For L

It shld start from the right

[wild linden]

,w {{1, 0, 0},{0,1,0},{0,-2,1}}.{{1,0,0},{0,1,0},{2,0,1}}.{{1, 0, 0},{2,1,0},{0,0,1}}.

[rocky lotus]

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[wild linden]

I don't think that's correct either

Bottom left entry must be a 4

,w matrix inverse {{1, 0, 0},{2,1,0},{0,0,1}}

[rocky lotus]

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[devout holly]

Hmmmmmm

But the U isn't wrong tho?

Can't tell

[wild linden]

Your three row operations seem to have been done correctly.

,w {{1,0,0},{-2,1,0},{0,0,1}}.{{1,1,2,4},{2,0,1,1},{4,4,2,5}}

,w {{1,0,0},{0,1,0},{0,-2,1}}.{{1,1,2,4},{2,0,1,1},{4,4,2,5}}

[rocky lotus]

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[wild linden]

,w {{1,0,0},{-2,1,0},{0,0,1}}.{{1,0,0},{0,1,0},{0,-2,1}}.{{1,1,2,4},{2,0,1,1},{4,4,2,5}}

[rocky lotus]

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[wild linden]

,w {{1,0,0},{0,1,0},{0,2,1}}.{{1,0,0},{-2,1,0},{0,0,1}}.{{1,0,0},{0,1,0},{0,-2,1}}.{{1,1,2,4},{2,0,1,1},{4,4,2,5}}

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[wild linden]

@devout holly you had your row operations wrong

,w {{1,0,0},{0,1,0},{0,2,1}}.{{1,0,0},{2,1,0},{0,0,1}}.{{1,0,0},{0,1,0},{0,-2,1}}

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[wild linden]

That should be your L matrix

[vale dock]

@devout holly Has your question been resolved?

[vale dock]

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[lone furnace]

have you tried polar form

wait no

[glass kelp]

seems factorable

is this to be considered geometrically?

[cloud epoch]

Do you have any ideas where to begin even?

have you tried anything?

[glass kelp]

or its a complex analysis q. in which case byeeeee

[cloud epoch]

You can do what you want, all you need is an upper bound

[silent dune]

Might be useful to consider z = 2e^(ix)

[quiet ledge]

it's complex analysis

[lone furnace]

what is maximum modulus principle again i forgot everything abt cmplex numbers

[silent dune]

This is on the boundary so not helpful

It says a holomorphic fucntion attains max at boundary

[lone furnace]

im generally not any helpful if it concerns complex numbers lol

[cloud epoch]

all you need is to use triangle inequality, like always in these problems

Once you have factored like this

[novel canyon]

complex analysis 🥀

[cloud epoch]

although that's not a correct factorization k

i just noticed

[silent dune]

What’s the fuzz about saying complex analysis

[cloud epoch]

should be |z^2 - 2||z^2 - 3|

[silent dune]

It’s easier than real analysis, and very elegant

[quiet ledge]

it can be also factorised to |z^2 -3||z^2-2|

[cloud epoch]

that's what i wrote two messages above

[lone furnace]

no it isnt lol

[flint creek]

No way

[cloud epoch]

His factorization was false

[glass kelp]

yes this

mine was z^4. -5z^2 - 6 🥀

shi

[silent dune]

I took complex analysis before real analysis

[flint creek]

You can't factorise two ways

[lone furnace]

you have a messed up syllabus

[cloud epoch]

can yall discuss your preferences in math elsewhere so we can help aditya with his problem please??

[silent dune]

I made the choice

[cloud epoch]

thank you

[flint creek]

One is right and other is wrong

[cloud epoch]

stop clogging the channel plz

[quiet ledge]

her? 💀

[lone furnace]

seriously, residue theorem is easier than heine borel bro???

[cloud epoch]

lol

my bad

[flint creek]

You think aditya is a girl???

[quiet ledge]

what are u? "it"?

[cloud epoch]

maybe

[quiet ledge]

ok jokes apart

[cloud epoch]

who knows

so do you know where to start aditya?

after factorizing

[lone furnace]

consider a geometrical approach

[silent dune]

.

[lone furnace]

at least, i guess

[cloud epoch]

No you dont need all that

[lone furnace]

thats my first sentence

[quiet ledge]

triangle inequality

[cloud epoch]

yeah

that should work

so what do you get then?

[quiet ledge]

it didn't i am getting multiple answers

[cloud epoch]

there are multiple

[silent dune]

Oh it says find an upper bound lmao

[cloud epoch]

there's an infinite amount of upper bounds that work

if x works, x+1 too

so if you find something, its enough

[quiet ledge]

no , it has to be single answer. according to book (it seems)

[cloud epoch]

might be the best answer they are asking for then

[silent dune]

Then the question is ill formed

[cloud epoch]

Yeah

[glass kelp]

maybe i tis the least upper bound?

[silent dune]

Unless the upper bound is infty lmao

[cloud epoch]

yeah, the best one

[orchid otter]

just use triangle inequality you should get single upper bound

[quiet ledge]

i used inequality but there's more to it, it seems

i am getting 1/2, 1/10, 1 and infinity

[glass kelp]

just wondering chat

im learning it w/ aditya

[quiet ledge]

i need to eat food, i'll be back. hopefully someone finds a solution

[orchid otter]

just use |z^4-5z^2+6| <= |z^4|+|5z^2|+6

[glass kelp]

|z^2 - 3||z^2 - 2| <= (|z^2| + 3)(|z^2| + 2) = |z^2|^2 + 5|z^2| + 6 <= 16 + 20 + 6 = 32, right?

[glass kelp]

oh

thx

[quiet ledge]

I think that will give lower bound instead of upper for the reciprocal

[glass kelp]

that gives u the least upper bound

[glass kelp]

why

[quiet ledge]

Reciprocating both sides inverses inequality

No?

[glass kelp]

oh right

🤔

upper bound of
[ \frac{1}{|z^4 - 5z^2 + 6|} = \frac{1}{|z^2 - 3|}\frac{1}{|z^2 - 2|}]?

nahh

[rocky lotus]

k

[glass kelp]

wait

nah

[orchid otter]

also didnt notice
|z^4-5z^2+6| >= min( | ±|z^4|±|5z^2|±6 |)

now i understand your issue
in this case you can use any combinations of signs, but you need to understand that value that you will get is bound but not necessary is exact

nvm what i wote above seems incorrect

[vale dock]

@quiet ledge Has your question been resolved?

[wicked heart]

upper bound means maxima?

[lyric bloom]

Can we let z^2 = t

To make it cleaner?

|t| = 4

[orchid otter]

we can
|z^2-3| >= |z^2|-3 = 1
|z^2-2| >= |z^2|-2 = 2
ensuring that we have eq on z=2
|z^4 - 5z^2 + 6| >= 2 with eq on z=2

[quiet ledge]

I don't think so. It is supposed to be limited by triangle inequalities

[quiet ledge]

Oh, hmm this looks good.. uhh lemme check

Idk why I got 49

[orchid otter]

upper bound indeed maxima but we need find it for reciprocal so we actually looking for minima of |z^4-5z^2+6|

[quiet ledge]

so answer is 1/2?

okay okay umm

yeah that's good

but uh
|z^4-5z^2+6| >= ||z^4| - |5z^2-6||
14 <=|5z^2 - 6|<= 26
i get
|z^4-5z^2+6| >= 2 and/or idk |z^4-5z^2+6| >= 10
that means answer is 1/10 or 1/2
also if take 14 <=|5z^2 - 6|<= 26 = 16 then answer is 1/0 = infinity

@orchid otter

[orchid otter]

here you need to avoid applying
|5z^2 - 6| = some value, because of that you fixed z value

you can think about it in this way with |5z^2 - 6| =16 you get
|z^4-5z^2+6| >= 0 but so 0 is lower bound but not necessary exact lower bound

also when |5z^2 - 6| = 26
|z^4-5z^2+6| >= 10, but not necessary exact lower bound and in this case
its only for z values that |5z^2 - 6| = 26 (z^2=-4 for example)

[orchid otter]

because of similar thing we ensure that we have z that makes eq both here, so this is exact bound

[quiet ledge]

i don't think i have fixing z with |5z^2-6| = 16 because, z is two variable x,y .
anyways, 16 is a possible value.
14 <=|5z^2 - 6|<= 26 this thing stands true for anything. right.. so
uhh idk

i think answer is either 1/10 or inf

[orchid otter]

okay lets say we have person that walk on stairs from 1st floor to second
his head always above his feet H_h > H_f
but we know that his feet somewhere between 1st and 2nd floor
this doesnt mean that his head is always above 2nd floor

[quiet ledge]

do u mean "this doesnt mean that his head is always between 1st and 2nd floor"

[orchid otter]

i mean if H_h > H_f and 1st_floor<H_f<2nd_floor
this doesnt mean that H_h>2nd_floor always true

[quiet ledge]

his head can never be less than 1st floor as long as 1=<H_f<=2
H_h can be >= 2 if 1=<H_f<=2 (let's say 1.7)

hmm

i can't see how this anology applies to our problem.

[orchid otter]

it can be >=2 but you cant say that its always >=2
so when |5z^2 - 6| = 26 you get |z^4-5z^2+6| >= 10 but its not for all z

[quiet ledge]

ok so you are saying is, we don't know value of |5z2 - 6|. we only know its whereabouts

but for all z , |5z^2 - 6| must lie between 14 and 26 right?

[orchid otter]

|z^4-5z^2+6| >= ||z^4| - |5z^2-6||
14 <=|5z^2 - 6|<= 26

correctly say |z^4-5z^2+6| >= ||z^4| - T|
for any z this is correct for some T in range [14, 26]
but not necessery for any in that range

[quiet ledge]

but that means 14 <=|5z^2 - 6|<= 26 is not true inequality?

[orchid otter]

its true you just cant apply it this way

returning back to analogy
1<=H_f<=2 is value with known interval
and H_h > H_f
its incorrectly to say that H_h>2 always

[quiet ledge]

look at this, they did nested inequalities. but i don't understand
one thing i understand is. they took 5z-1 = 11 because 11 is closer to 16 to make denominator smallest possible

[slim summit]

We haven't already done this exercise together a few days ago 🧐

Oh ok not the Same

Almost xd

[quiet ledge]

actually, 1+H_f<= H_h <=2+H_f

[orchid otter]

in this case it worked only because |z^4| > |5z-1| so to minimize
|z^4| - |5z-1| we maximize |5z-1|

[orchid otter]

yes you can see that H_h and H_f are related
in other words not any H_h(t1)>H_f(t2) where t1, t2 are different time moments

[quiet ledge]

how do we know we need to maximize |5z-1| to minimize |z^4| - |5z-1|?
|z ^4| is exponential it will increase faster |5z-1| right? so
increasing |5z-1| will also increase |z^4| even greater ?

[orchid otter]

we got fixed |z|

[quiet ledge]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

so |z^4| is also fixed

[orchid otter]

basically this will work when you have one value that bigger than all others together, it will not work in problem with 5z^2

[quiet ledge]

so to minimize ||z^4| - |5z-1|| we need |5z-1| closest to |z^4| = 16

so we had 9<=|5z-1| <=11 so we take 11 which is closer to 16
but what we can do when for example: 10<= ......<=20 ?

[orchid otter]

yes here it works, if we try to apply it in prev problem with 5z^2 we will be able to get 0 as lower bound but cant guarantee that its exact boud

[orchid otter]

yeah in this case you also get lower bound, but you at this point also dont know if its exact bound

so you need provide some z value to prove that its exact

[quiet ledge]

this part i am not able to understand that "we dont know its exact bound"

[orchid otter]

because we have restrictions in this case z^4 and 5z^2-6 are not just any values with such modulo,they have relation

its not exactly the same but if you have sides with length 3, 5 the third side of triangle is bigger than 2, but lets say we have restriction, for example angle between 3 and 5 sides, lets say its like >=30 , then while triangular inequality says that third side >=2 but we cant get 2

[lyric bloom]

Did you guys get the answer?

[quiet ledge]

also if u factorise |z^2-3| = |z+3||z-3| and |z^2-2| = |z+2||z-2| we get |z^4 - 5z^2 + 6|>= (3-2)(3-2)(2-2)(2-2) = 0

[orchid otter]

u forgot sqr root
|z^4 - 5z^2 + 6|>=0 is not breaking anything

its correct lower bound just not exact

but we need exact

[quiet ledge]

then how we know 1/2 is exact?

[quiet ledge]

why sqr root?

[orchid otter]

z = 2

|z^2-3| = |z+sqrt(3)||z-sqrt(3)|

[quiet ledge]

ohhhhh yeah my bad

[orchid otter]

we get eq with z = 2, you can also ensure that all ineq in process also become eq with z=2

[quiet ledge]

wdym

z = 2 like z = x+iy = 2?

[orchid otter]

yeah if you just calculate for this z you will get 1/2, this way we show that this is exact bound

in simple cases
problems with single inequality are usually simple with this
there is usually one point when ineq is eq

but when you use different ineq sometimes then become eq on different cases, in this case its become dirty

in case
|z^2-3| >= |z^2|-3 = 1
|z^2-2| >= |z^2|-2 = 2
ensuring that we have eq on z=2

we actually got lucky that they have same point of eq (actually its problem design but anyway)

[quiet ledge]

wdym eq?

[covert gazelle]

is this the question

[quiet ledge]

yes

[orchid otter]

equality, so i meant z value when ineq reachs eq

[quiet ledge]

thanks @orchid otter i am just getting headache thinking more about this.
i just want to be able to solve all kinds of problem. i wish i had question bank realated to triangle inequalities.
i still don't understand completely

[orchid otter]

its fine you just need to remember that if you get some boundary it could be correct but not exact

[quiet ledge]

yeah just like -infinity to infinity correct for all real numbers and expressions but not exact for equations or set of equations and inequalities?

[orchid otter]

yeah, that why in such tasks we need provide values that will reach boundary that we got

only bad thing is you cant know if boundary is incorrect, you can know only if its correct if you find values to reach that boundary

[quiet ledge]

hmmm, that makes sense

i wish we were taught some general method to solve this kind of stuff

anyways

thanks

thanks a lot

🫡

[vale dock]

@quiet ledge Has your question been resolved?

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[civic hill]

why are my answers wrong

[stark wedge]

well for starters you put latin a instead of greek alpha everywhere

[glass kelp]

alpha and a

${\alpha}$ and ${a}$

[rocky lotus]

k

[glass kelp]

💔

[stark wedge]

but also i am p sure the third answer is wrong even post-correction, so double-check that!

[civic hill]

vro how am I suppose to tell the difference if it is so tiny

[stark wedge]

please do not call me "bro"/"brother" or any variations thereof!

could you edit that word out, please?

[civic hill]

I just say that to everyone lol

no matter who you are

[civic hill]

also this was right

[civic hill]

thanks

[stark wedge]

ngl this is a weak excuse

[civic hill]

.close

[vale dock]

Channel closed

Closed by @civic hill

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[balmy monolith]

how do you solve this without using newton's method or r alpha method (easier method) after getting sin(2x) + 2cos(2x) = -2

[wraith heart]

use pythagorean theorem then quadratic formula

[balmy monolith]

oh

[scarlet sandal]

Why does the r alpha method not appeal

[balmy monolith]

how exactly would you do that

[scarlet sandal]

seems the easiest

[balmy monolith]

we weren't taught that method in class

[delicate wagon]

Hello

Where are you guysfrom?

[scarlet sandal]

does that mean you can't use it?

[delicate wagon]

Country?

[west cloud]

hello, this channel is occupied; for general discussion please see #360643390594875392. if you have a specific math question, see #❓how-to-get-help

[balmy monolith]

well i used it anyway but i was hoping to find an easier version

[scarlet sandal]

I think r alpha is the most straightforward method

although I get that you want to see how it's done in class