precalculus

[gilded fern]

Anyone midn helping me with this

I used algebra to solve for each side

But I do not get the correct answer

Which is 7degrees

[gaunt mason]

@viscid thistle I don't understand your query.

A straight line is uniquely identified by 2 points. Meaning given some two points there is exactly one line passing through them.

The points and the straight line, and curved line, and circle and any other weird shape can in mathematics be put not just on simple empty paper, but also on the paper that has a coordinate system define on it.

With its center, and x and y axis that are perpendicular to each other.
The position of points in this coordinate system is given by (x,y), for example (3,0). The line is given as y=kx+b for example y=-2x + 3. The only exception is the line parallel to the y-axis.

[visual kayak]

@viscid thistle lol you have the correct answer. To find the slope you did it correctly but i don't know what the second half is about. The question gives you the y-int (0,5) so there shouldn't be anymore work you have to do if I am correct.

[gaunt mason]

@gilded fern What was your procedure?

[gilded fern]

I solved for:
-3theta + 76 = 90
and
-2theta + 49 = 90

[viscid thistle]

Thanks @visual kayak

[sick steppe]

@gilded fern you cant have tan(90)

[runic sundial]

So I'm developing a first person game, and I want to make realistic player movement. To move the player, all I have to do is call one line of code AddForce(), in which I pass in a value that acts as the amount of force being applied. The problem with this is that if I add a constant amount of force, then the player takes a really long time to accelerate and it moves really fast, so what I have to do is change how much force is being applied based on the velocity of the player and the drag and gravity but I don't know exactly how to calculate that. Is there maybe a function similar to F=MA that I can use that takes drag and gravity into account? Thank you!

[gaunt mason]

I dont understand, the force applied to the player is independent of the drag and gravity

But the friction might give an opposite force @runic sundial

[runic sundial]

No I know it's independent

but it's a game engine so it doesn't really simulate drag very well

at least not in my experience

And I just realized I meant to say friction instead of drag

[gaunt mason]

The friction is a constant force, so it applies a constant acceleration.

[runic sundial]

Okay well then is there a way to calculate the amount of force I need to apply to maintain a constant speed

[gaunt mason]

The friction force is F=mgu where u is the friction coefficient

This is the amount of force you have to apply to maintain the constant speed.

[sick steppe]

That's if the normal force = gravity force

$F_f = \mu \cdot F_N$

[obsidian monolith]

moshill1:

[ancient rampart]

could i get help with this

[sick steppe]

@ancient rampart set up a triangle likely

[ancient rampart]

hm

[sick steppe]

Either quad 2 or quad 4

[ancient rampart]

def quad 4 cuz inverse tan

[sick steppe]

Tan is neg in quad 2 or 4

quad 4 will just make the final answer pos

[ancient rampart]

i think my teacher said smth abt if the inverse is given in the problem you use the "universally accepted" restricted domain

which for tan is -pi/2 < pheta < pi/2

[sick steppe]

yeah but arctan is defined for all values

[ancient rampart]

ill just do it in q4

[sick steppe]

oh yeah true woop

so yeah, do 4

[ancient rampart]

and then cos is x/r so -4/5?

[sick steppe]

radius is never negative

[ancient rampart]

oh

oh yea

oops

so 4/5

[sick steppe]

ye

[granite crescent]

can i please get help with #9?

[sick steppe]

extrema occur when dy/dx = 0

[granite crescent]

so far i was able to find this

this is the derivative of the function

[sick steppe]

you can get a quadratic in sinx and solve that

[granite crescent]

using the trig stuff ?

[sick steppe]

be more specific

than "trig stuff"

[granite crescent]

trig identities

[patent beacon]

@runic sundial
Consider instead a speed cap, which would be much more like real life.

[sick steppe]

yes

[granite crescent]

does the 0<x<2 pi change?

[sick steppe]

no

[granite crescent]

why?

[sick steppe]

cause that's just the interval you're looking at for y

[granite crescent]

i got this rn

[sick steppe]

yeah

[granite crescent]

we’d use the quadratic formula

but it doesn’t work since the negative sign is in

[sick steppe]

yep

[granite crescent]

so what should i do

[sick steppe]

write the conclusion statement about the maxes and mins on the interval

[granite crescent]

but there are none

[sick steppe]

yeah

[granite crescent]

oh

[sick steppe]

You can have no maxes and no mins on an interval

If 0 and 2pi were included in the interval then you'd have some

[ripe adder]

Im trying to find the length and width of a rectangle inscribed in a circle

so far the online resources I have looked at have only made it more confusing

[fading token]

Hint: write down two equations involving both length and width of the cut-out

[ripe adder]

Yes I get that

but im confused on how to get those length and width equasions

I know that P = 2W=2L

and A = W*L

but I would need perimeter to find the width and length with systems of equasions no?

[fading token]

You need the area equation, but you do not need the perimeter equation

There's another one that can help you

[ripe adder]

I also know that 20^2 = L^2 + W^2

[fading token]

Precisely

There are your two equations

So you can solve for both L and W

[ripe adder]

@fading token this doesnt look right to me

[fading token]

Why not?

Apart from the fact that you shouldn't use the negative values for L and W for obvious reasons

[ripe adder]

oh shi

ur right

sorry most my answers are usually whole numbers or fractions

thank you for the help

[fading token]

np

[drowsy karma]

i know this shit easy, i just dunno how to do it

@fading token could u help me real quick

[calm hornet]

@drowsy karma first find the r (multiplyer) of the sequence

r = u2/u1 = 12/6sqrt(2) = sqrt(2)

[drowsy karma]

the difference of squares?

orrr

[calm hornet]

r = u2/u1 = 12/6sqrt(2) = sqrt(2)
do you understand this one?

[drowsy karma]

Not really. He didn't teach us this.

[calm hornet]

ahh

[drowsy karma]

I've done like dividing/multiplying radicals n stuff

[calm hornet]

first or all you need to understand what is geometric sequence

ex: 3, 6, 12, 24, 48, 96...
7, 21, 63, 129, 387,...

a geometric sequence has a common multiplyer r

[drowsy karma]

Okay, so you're finding the common multiplier in these two terms?

[calm hornet]

yes

from your question

the first term is 6sqrt(2)

[drowsy karma]

yeah

[calm hornet]

and the second term is 12

so the common multiplyer is sqrt2

which means the third term should be 12sqrt2

get it?

[drowsy karma]

let me process this

okay yes i think so

I just had to put it in radical form

now I get it

12 was 144sqrt, 6sqrt(2) was 72

oHHhh

now its 288

[runic sundial]

That doesn't really work

I tried it and the movement doesn't feel natural at all

[viscid thistle]

Can someone show me their steps for the first problem

I am brain blocked right now

[sour hemlock]

just factorize

[viscid thistle]

My brain is blocked

I know how to factorize

[sour hemlock]

if you are not feeling well, I suggest you get some rest

and then try again

[viscid thistle]

It's just not on it ya know

I'm falling behind, if you can just show me the steps for just the first one that would be really helpful

[sour hemlock]

let y = 0 and then factorize

[loud forge]

Can someone help me with this

[willow bear]

what's giving you trouble here?

[sour hemlock]

I guess you should first start by drawing the triangle and labeling the sides

then it will be clear to you

[willow bear]

i mean, heron's formula can work

[sour hemlock]

damn I never used that, I would just draw the triangle and draw the perpendicular and find its height haha

[cyan hill]

Searched up what herons formula was

It feels like cheating ngl

[loud forge]

Ima use that thanks

[patent beacon]

What's wrong with it? Haha

[runic sundial]

@patent beacon It's being kind of jittery

as if I was using a character controller if you know what that is

[patent beacon]

No I don't oop

[runic sundial]

Okay well it doesn't work very well

I have a stack overflow thread for this question: https://stackoverflow.com/q/64978310/9362751

[patent beacon]

@runic sundial
An object moves at constant velocity if the net force on it is zero

[runic sundial]

If friction isn't involved

But friction is involved

[patent beacon]

Friction is a force

Anyway so an object moves at constant velocity if the net force on it is zero

That is, all of the forces on that body have to cancel out

So if you do have 10N of friction pulling on an object, then you'd need 10N of applied force to keep it moving

[runic sundial]

I'm sorry but i don't really know how to apply that into code

@patent beacon You're telling me that friction is a force but unity already has friction built into it

But I still need to find a mathematical way to apply less force over time to retain a velocity

[patent beacon]

As soon as you have the speed you want, drop the applied force to be equal to friction

As far as unity programming goes, I'm no expert haha. But that's how it's done in real life

[runic sundial]

Ahhh no that makes sense

Wait one more thing

If I wanted to take into consideration friction AND gravity, would i just drop the speed to the friction + gravity or something instead of just friction?

or like friction * gravity

[patent beacon]

I'm not sure how unity treats these things oop

Usually the force of friction is increased by higher gravity

[runic sundial]

so then it would probably be friction * gravity

That helps out a lot! Thanks!

[patent beacon]

In real life,
Max friction force = (friction constant) (mass) (gravity)

That's if you are not going up or down an incline though

[loud forge]

Can someone help me with this, I don't even know where to get started

[willow bear]

are you familiar with half-angle identities

[loud forge]

No

Im gonna watch a youtube vid on it

Hopefully that'll help

[viscid thistle]

common denominator

use the fact that r!=r(r-1)!

[jagged glade]

r! = r(r-1)(r-2)...(3)(2)(1) = r(r-1)!

(r-1)! = (r-1)(r-2)(r-3)...(3)(2)(1)

[viscid thistle]

do you what a factorial is

[jagged glade]

,w factorial

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Biscuit#5208

[jagged glade]

https://www.mathsisfun.com/numbers/factorial.html

[viscid thistle]

all that spam is not necessary lol

k! = k*(k-1)!

[opaque olive]

could someone help me on this one 😩

<@&286206848099549185>

[velvet blade]

Why is complex number z= a + bi have an inverse w such that $w = \frac {\bar{z}}{a^2 + b^2}$ and not just $ \frac {1}{a+bi}$?

[obsidian monolith]

lazypawtato:

[opaque olive]

https://media.discordapp.net/attachments/363224154469826562/780746710606413844/unknown.png

halp plez

i dont know the point P which is stopping me from answering this :/

[viscid thistle]

it's not inverse but reciprocal

it's easier if you have nothing complex in the denominator

so you are easily able to distinguish real and imaginary part

[opaque olive]

@viscid thistle could you have a look at my question please?

[velvet blade]

there's a difference?

[viscid thistle]

reciprocal is more precise

[velvet blade]

I didn't get you

precise how?

[viscid thistle]

we already talked about this

it's called reciprocal.

[velvet blade]

@opaque olive I'm not able to understand how to solve your qustion, does it need calculus or something?

[opaque olive]

well yeah -_-

[velvet blade]

this is precalc, ask in #calculus

[opaque olive]

whats precalc

[velvet blade]

everything before learning calcculus

[opaque olive]

ok

[velvet blade]

you said because it's easier to not have imaginary partr in denominator, is that why reciprocal is different from inverse?

[viscid thistle]

no

$ \frac {1}{a+bi}$ and $\frac {\bar{z}}{a^2 + b^2}$

[obsidian monolith]

HoboSas:

[velvet blade]

i'm sorry if my questions are annoying, but this is the first time i'm hearing about this difference

[viscid thistle]

what is $ \Re\left(\frac {1}{a+bi}\right)$?

[obsidian monolith]

HoboSas:

[viscid thistle]

it's hard to tell

while $ \Re\left(\frac {\bar{z}}{a^2 + b^2}\right)$ is easy

[obsidian monolith]

HoboSas:

[velvet blade]

Re?

[viscid thistle]

real part

[velvet blade]

Oh okay

[blissful kayak]

$\mathfrak{R}\left(\frac{\bar{\widetilde{z}}}{a^2+a^2}\right)=\frac{a}{a^2+b^2}$

[obsidian monolith]

Coleculus:

[blissful kayak]

Hence, $\mathfrak{R}(\widetilde{w})=\frac{x}{x^2+y^2}$

[obsidian monolith]

Coleculus:

[blissful kayak]

Is that right?

[small peak]

Hey, can someone teach me calculus for the IB math AI SL.
Im struggling pretty bad and have a test coming up

[sick steppe]

@small peak teach you all of calculus...?

[small peak]

not all of calculus

Chapter 12 : Analyzing rates of change - Differential Calculus

Chapter 13 : Approximating Irregular Spaces - Integration

just these @sick steppe

[quaint mason]

idk how to solve for 2nd value for this problem

for first one i just took theta=sin^-1(-.66)

[echo wagon]

In what other quadrant is sin negative?

[sick steppe]

@quaint mason well for starts that's cosine not sine

[viscid thistle]

?

[sick steppe]

oh wait

i cant read nvm

[quaint mason]

ik but im using the cofunction identity

[sick steppe]

yeah i realize that now lol

[quaint mason]

cos(pi/2-theta)=sin(theta)

[sick steppe]

yeah

[quaint mason]

after i found the value of theta, i then di pi-theta

which gave me 3.8624 for first value

[lunar axle]

can someone help me with these questions please??

[loud forge]

How do you do this?

[lapis sphinx]

set up a right triangle

with angle theta in quadrant 1

@loud forge

[loud forge]

Ty

[lapis sphinx]

np

[quaint mason]

how do i solve for 2nd value of this?

[sick steppe]

@quaint mason sine is neg in quad 3 and 4

quad 3 is pi + principle angle, quad 4 is 2pi - principle angle

[quaint mason]

so the first value is pi-sin^-1(-.66), the second would be 2pi-sin^-1(-.66)?

whats principle angle

[sick steppe]

$\alpha = |\sin^{-1}{(-0.66)}|$

[obsidian monolith]

moshill1:

[sick steppe]

$\theta = \pi + \alpha \text{and} 2\pi - \alpha$

[quaint mason]

huh

[obsidian monolith]

moshill1:

[quaint mason]

thats what i did n it was wrong

[sick steppe]

pi + alpha is the 3.8 angle

2pi - alpha ~ 5.56

[quaint mason]

oh ik what i did wrong w the 2nd value

i rounded it wrong

thank u

also, quick question, what does this symbol mean

β

is it just like x?

im just tryna simplify tan(-β)cos(-β) which is= -tan(β)cos(β) but just wanted to make sure if β meant somethin important

[patent beacon]

It's just a placeholder

tan(-🍞)cos(-🍞) = -tan(🍞)cos(🍞)

[quaint mason]

ah

[sick steppe]

$\beta$

[obsidian monolith]

moshill1:

[lapis sphinx]

$\gamma$

[obsidian monolith]

North:

[lapis sphinx]

$\omega$

[obsidian monolith]

North:

[viscid thistle]

#bots

[ripe adder]

im trying to solve this fraction decomposition problem

and I dont know how to do it

due to the third variable

[ripe adder]

<@&286206848099549185>

[sour hemlock]

why don't you multiply both sides with the denominator of the LHS

you will get an equation no longer in fraction form

then make appropriate substitutions for x to solve for A, B, and C

[ripe adder]

Ok, how about a new problem

[quaint mason]

i was able to figure out one of radian value is 1.37, how would i solve for the other ones?

[zealous parcel]

2sintheta - 3csctheta = 2

i have to solve for pi radians within the unit circle without using a calculator (using factoring instead)

[velvet blade]

I got the answer as 1, But I'm not sure of my method

[viscid thistle]

show work

[velvet blade]

is |z|^2=z^2 for a complex number?

[viscid thistle]

no

|z|^2=a^2+b^2

z^2=a^2-b^2

[velvet blade]

Ohh

damn

Can you show the correct way?

wait, it's 1/|z|^2 right?

wait , |a||b| = |ab| right?

[viscid thistle]

yes

[velvet blade]

okay I also need help proving triangle inequality then

[viscid thistle]

just square both sides

[velvet blade]

Oh okay, got it.

So what do I do about this absolute value of z/conj(z)?

[viscid thistle]

try to multiply by conj/conj

[velvet blade]

|z|^2/conj(z)?

that's it?

wait why to do all that?

$|\frac{z}{\bar{z}}| = \frac{|z|}{|\bar{z}|}$ right?

[obsidian monolith]

lazypawtato:

[velvet blade]

@viscid thistle you there?

[viscid thistle]

not really

[velvet blade]

Bruh

[sour plinth]

you could sub in z=a+ib

[viscid thistle]

that is true

[velvet blade]

Okay so it |conj(z)| = sqrt(a^2+(-b)^2)

=|z|

so it's 1?

[sour plinth]

yes

[velvet blade]

thanks man

[velvet blade]

did he mean value of z/r is 1 or |z/r| is 1?

[past meadow]

researcher |z/r|=1 is what he meant

[velvet blade]

thanks

[ripe adder]

what steps should I take to find a / b?

[willow bear]

do the words "partial fraction decomposition" ring any bells to you

alternatively: have you taken the time to follow the hint?

[ripe adder]

following juans hint I got that

a+b = 1/x

did you send me something I didnt see?

[willow bear]

a+b = 1/x
how?

can you show your work?

that looks really weird

[ripe adder]

((a/(x+1))+(b/(x-1))=1/(x^2-1)(x+1)

=

a+b=1/x

I did this because i was multiplying by LHS denominator

like juan said

[willow bear]

uh

okay wai

do you have your work ON PAPER

you tried to multiply both sides by... x+1?

[willow bear]

that'd give you $a + \frac{b(x+1)}{x-1} = \frac{1}{x-1}$, at best.

[obsidian monolith]

Ann:

[willow bear]

not a + b on the left hand side, and definitely not 1/x on the right!

[ripe adder]

ah I see

wouldnt the right hand side be

1(x+1)/x^2-1

or

wait

nvm

i see

ok ill balance that

[willow bear]

you're going down a road that is, at best, a massive detour

[ripe adder]

so then how do I solve for a and b?

[willow bear]

why would you need the ratio of a and b

anyway, you might simplify the left-hand side,

as the hint told you

so combine the two fractions into one

so $\frac{a}{x+1} + \frac{b}{x-1}$ simplifies into: $\frac{a(x-1) + b(x+1)}{(x+1)(x-1)}$

[obsidian monolith]

Ann:

[willow bear]

understand? y/n

[ripe adder]

yes

multiplying the numerators / denoms to add them

[gentle fox]

is this the right place to post questions on probability?

[willow bear]

@gentle fox #probability-statistics

and the numerator, once you expand & collect like terms, becomes: $$(a+b)x + (-a + b)$$

[obsidian monolith]

Ann:

[willow bear]

so you have: $\frac{(a+b)x + (-a+b)}{x^2 - 1} = \frac{1}{x^2 - 1}$ \ \ and thus, $(a+b)x + (-a+b)$ has to be equal to $0x + 1$ for all values of $x$

[obsidian monolith]

Ann:

[willow bear]

thus: $a+b = 0$ and $-a + b = 1$

[obsidian monolith]

Ann:

[ripe adder]

so

I am on the first of those four

I simplified

into (ax-a+bx+b)/(x^2-1)

how did you collect the like terms to reorganize it like that

the one I replied to with ghost emoji

[willow bear]

distributive property

ax + bx = (a+b)x

[ripe adder]

I see

ok so

knowing this information

how do i progress back towards the original format

of

I see

a = -1/2

b = 1/2

yes

thank you @willow bear

[velvet blade]

I understood the question but I didn't understand the answer

Also is e here euler's number or just a notation?

Because he didn't teach anything about why e is used in the notation

[crimson steppe]

it's euler's number from what i recall

[astral mantle]

it is

exactly that

[crimson steppe]

yup lmao nice to see you again lol

[astral mantle]

it talks about the euler identity in the example

[willow bear]

Also is e here euler's number or just a notation?
yeah, it is euler's number

you're getting a glimpse of the wonderful world that is the complex exponential

i mean, $(e^{\frac{i\pi}{2n}})^n = e^{\frac{i\pi}{2n} \cdot n} = e^{i\pi/2}$ so...

[obsidian monolith]

Ann:

[willow bear]

¯_(ツ)_/¯

[crimson steppe]

yeah, that magic number lol

it just does shit and it works

[willow bear]

eh

i mean yeah i guess it kinda takes a lot of math to explain with any amount of rigor WHY $e^{ix} = \cos(x)+i\sin(x)$

[obsidian monolith]

Ann:

[crimson steppe]

yeah, that's what i read in the book

it's still a really nice shorthand

for the polar form lol

[lunar axle]

help please- with either one or both if possible

[hexed granite]

if anyone can answer any of these it would be a life saver

please someone help

<@&286206848099549185> please

[obsidian monolith]

lazypawtato:

[willow bear]

$\frac{z}{r} = 1(\cos(\theta) + i \sin(\theta))$ if you so insist

[obsidian monolith]

Ann:

[velvet blade]

Yeah but the book says only absolute of the fraction is equal to 1 right?

[willow bear]

and when did i say otherwise

[velvet blade]

but z/r isn't absolute right?

[willow bear]

i didn't require theta to be 0, did i?

haven't you already seen that every complex number can be written as the product of its magnitude (absolute value, modulus, whatever you wanna call it) and a complex number on the unit circle?

[velvet blade]

...... I'm confused, it's just that the book said you can have any real number theta such that cos(theta)+isin(theta) when |z/r| is 1. I don't understand how you don't take absolute value of z/r and still use cos(theta)+i sin (theta)

[willow bear]

...

did you misread the book or sth

[velvet blade]

I feel so

give me a min

It says |z/r| is 1

so the solutions exist on the circle of radius 1 right?

[viscid thistle]

@lunar axle are you still stuck? If so come to another channel and tag me there.

[willow bear]

so the solutions exist on the circle of radius 1 right?
overthinking alert

i mean yeah z/r is on the unit circle

since

yknow

that's where the points with magnitude 1 are

[velvet blade]

I'm unable to put it in words what I understand

Wait! I think I got it

|z/r| = sqrt {cos^2(a)+sin^2(a)}

Am I right?

[willow bear]

eh

i guess?? technically

[velvet blade]

bruh

I'll try again tomorrow I guess

Thanks

BTW why do we use exponent base as e for writing it in polar form?

Do you learn it in calculus?

Is e a variable?

in this?

[willow bear]

no, it's a constant

it's euler's number

2.71828ish

[velvet blade]

I thought you hated decimals

[willow bear]

i said ish

it's irrational lmao

[velvet blade]

Okay

[willow bear]

transcendental even

[velvet blade]

what's transcendental?

[willow bear]

not expressible as the solution of a polynomial equation with integer coefficients

[velvet blade]

Oh So use numerical analysis to approximate it?

[willow bear]

if approximating the value of e itself is your goal then yeah i guess ¯_(ツ)_/¯

[velvet blade]

Okay, why do you use that as the base though? You're not even using it to convert it in polar form back again

[willow bear]

i mean

because $e^{ix} = \cos(x) + i \sin(x)$ only works when the base is $e$ lol

[obsidian monolith]

Ann:

[velvet blade]

Oh, do you learn in calculus about why it works?

[willow bear]

yes

maybe in the later stages of calculus

definitely not beginner stuff, that

[velvet blade]

Nicee!

Thanks a lot Ann

[potent berry]

can someone help please? been trying to figure it out myself and with internet tool but couldnt

[sour hemlock]

well you can multiply

[trail niche]

You’re solving for A I assume?

Just remember that i^2 = -1

And distribute

[potent berry]

i tried to multiply using the rules (a+b)(a-b) and so on, knowing i^2 = -1 and i still couldnt get the right answer

yes solving for a

[trail niche]

Paste the work you did

[potent berry]

sorry for my handwriting lol

my answer is incorrect ofc

[trail niche]

Yeah

You forgot a negative sign

Try +1/4

[potent berry]

where did i forget?

a is supposed to be -1

[trail niche]

Wait

Give me 30 secs

[potent berry]

no pressure 🙂

[trail niche]

Also in your typed out question

You missed a ()

For where a-2 belongs to

I’ll assume it’s just a 3rd term tho

Same thing

Nice

My pencil just broke

And all the lead is stuck inside

BRB

Ok back

I got 1/2, and -1

@potent berry

Idk where you were going with the second half

But you were doing alright in the first half

[potent berry]

haha can u send a pic of your solution?

[trail niche]

You got A isolated to be (A^2 + 2A + 4)(A-2) = -9

You keep going there and you can jsut solve for A

[potent berry]

i would keep going if i knew how 🙂

[trail niche]

Well let me send you a pic

Give me. A sec

[potent berry]

ty

second half i used a calculator to solve (a^2+2A+4) but now i realise its stupid because the calculator assumes it equals 0

[trail niche]

Lol yeah

You gotta specify x

Or it’ll think A is a constant or something

[potent berry]

ayy

got it

no need for pic

[trail niche]

Hold on still

[potent berry]

thank you sir

[trail niche]

Oh ok

You’re welcome : D

You should have ended up with A^3 = -1

But there’s also x = 1/2 apparently

[potent berry]

yeah

theres also x=1/2 +i(root3)/2 and 1/2 - i(root3)/2

but theyre not needed or marked as solutions in my homework

so im fine with just -1 haha

if "a" was defined as complex i could find them tho

its defined as real in the question

[trail niche]

Ye

Alright nice

👍

[ripe adder]

So I set these as equal to eachother and then multiply the denominator?

[steep quartz]

Yes. what equation do you have after that?

[trail niche]

^

You (x-3)^2 is equal to x^2-6x+9

So you can multiply the bottom left fraction by (x-3) on top and bottom

So you get a common denominator on all 3 fractions

[sand gate]

Is this correct?

[bronze star]

Check your domain again

[trail niche]

^

@sand gate remember that your domain is your x values

So try and draw vertical lines at the extremes

You’ll. Realize that at one of them you can’t (cause it goes to -infinity) but at the other end yo can

[rancid crest]

given f(x) = x^3 + kx -15 and the remainder when f(x) is divided by x -3 is - 18 then what is the value of k?

i got k = -10

mm

what if

@viscid thistle

[velvet blade]

if $w^2 = z$ then $ \sqrt{z} = +/- w$, that's it right?

[obsidian monolith]

lazypawtato:

[velvet blade]

Also I can't figure out the later questions

<@&286206848099549185>

[blissful ridge]

I don't think you argument is enough

[velvet blade]

Oh

[proud raven]

i mean

[blissful ridge]

Let square root z=x+iy be another complex number a+ib

i.e √z=a+ib

z=(a+b)²

And then you can compare real and imaginery parts
You'll get a quadratic while solving for a or b

So you'd get two values for each a and b

Hence two roots

I guess this should work

[velvet blade]

Oh okay, what about complex number w^n=z having at most n roots?

[cedar cipher]

Use polar form

[proud raven]

thats my thinking

usually wrong

[cedar cipher]

It's -b^2,not -b^2i

[proud raven]

oh youre right am dumb

i hate complex numbers

[velvet blade]

Okay

[ebon fiber]

could someone help for solving log_2(n) >= log_12(n)? (edited

[velvet blade]

$\log_2{n} = x$ then $2^x = n$

[obsidian monolith]

lazypawtato:

[velvet blade]

z=(a+b)²
@blissful ridge it's (a+bi)^2 right?

Won't taking quadratic roots just give me +- (a+bi)?

[blissful ridge]

Ahh, yes

I missed the i

[velvet blade]

so what gives?

[blissful ridge]

What??

We are trying to find √z

[velvet blade]

yeah,that's +- w right?

[blissful ridge]

Our assumption was not z=(a+ib)²

It was √z=(a+ib)

We go from √z=(a+ib) to z=(a+ib)²

And then prove that it must have two roots

[velvet blade]

What's the difference?

I don't get it

[blissful ridge]

x²=4
And
x=√4

Is there a difference between the two??

[velvet blade]

x=+-2?

both give the same answer right?

[blissful ridge]

No

x²=4
Gives you two solutions ±2

But x=√4 only gives you one

x=2

[velvet blade]

I don't get why it's only a positive solution

[blissful ridge]

You cannot get a negative number out of square root

[velvet blade]

We do that in quadratic equations right?

[past meadow]

when we use the square root symbol on its own without a plus or minus we're specifically talking about the positive square root

[velvet blade]

Okay

How do I show there are two distinct complex numbers then? Just w^2 = z and (-w)^2 = z?

[velvet perch]

hello this is about exponential growth and decay

im just wondering and trying to distinguish between these qs

why does the first picture question use A (which is a constant representing the initial value)

but then the q in the 2nd pic just says it's A

if anyone can help me, that'd be great

[viscid thistle]

?

[velvet perch]

so you see that there is a constant A right

[viscid thistle]

yes I see that

[velvet perch]

i am wondering why it's applied to the first pic

but not the 2nd one

[viscid thistle]

what do you mean "applied" ?

[velvet perch]

like it's not given a value

[sick steppe]

There's an A in both pictures...?

[velvet perch]

if you get me

it's just A and it gets cancelled out

[viscid thistle]

I mean those are different exercise

you have different initial data

[sick steppe]

Typically the A cancels either way

[velvet perch]

so would air pressure just be 1 🤔

oh

i see

i guess that makes sense im just a bit slow

[sick steppe]

the e^kt part is just what percentage of A do you have after t time units

[velvet perch]

ohhh

oh

[sick steppe]

So if you want 10 percent of A, you'd have .1A=Ae^kt

-> .1 = e^kt

[velvet perch]

yeah yeah that makes so much more sense

thank you so much !!

[velvet blade]

Hey so I wanted to ask something about polynomials which was bothering me for a long time
Like not a problem, just a doubt about the property

[willow bear]

what is it

[viscid thistle]

yea ask

@velvet blade

[velvet blade]

@willow bear Sorry for delay, I remember asking you this already. But I wasn't convinced.

Give me a min to phrase my question. I had a small errand to do

[willow bear]

this again wow

[velvet blade]

If you remember this, I asked about the exceptional coefficients that exist right?

[willow bear]

the what now

[viscid thistle]

what do you want here?

[velvet blade]

Like he claims adding an n degree term with n-1 degree term doesn't change the degree

[willow bear]

yeah because n-1 is lower than n

[velvet blade]

What about the cases when the coefficients are the case such that it does?

[willow bear]

it doesnt

the only way the degreee of a polynomial can drop after adding something is if the leading terms of your polynomial and the thing you're adding cancel out

[viscid thistle]

guve an example where it does researcher

give*

[velvet blade]

Okay let's say we have $f(3) = 2*(3)^4+ 3(3)^3$

[willow bear]

this is not a polynomial it is a number

[velvet blade]

Sorry let me correct

[obsidian monolith]

lazypawtato:

[viscid thistle]

do you mean degree as in the sign?

[sick steppe]

degree = highest power

[velvet blade]

a_n=2, which is a coefficient

[willow bear]

no

[viscid thistle]

ok moshill1

[willow bear]

2 * 3^4 + 3 * 3^3 is a number

[viscid thistle]

how will the degree change

[willow bear]

considered as a polynomial this is a constant, specifically the constant 243, with degree zero

,calc 2 * 3^4 + 3 * 3^3

[obsidian monolith]

Result:

243

[velvet blade]

It's an output of the polynomial f(x) = 2x^4+ 3x^3 right?

[sick steppe]

yes

[velvet blade]

where x=3?

[willow bear]

well you didn't say your function was that

if you wanted to declare f(x) = 2x^4 + 3x^3 then you should have said f(x) = 2x^4 + 3x^3 instead of introducing it as this mystery function for which only the value at 3 is known!

[velvet blade]

okay sorry

[sick steppe]

it was implied the function was 2x^4 + 3x^3

[willow bear]

ok so what is it about the polynomial 2x^4 + 3x^3 that troubles you

[velvet blade]

That one input which changes the leading coefficient

as an example

[willow bear]

what the hell do you mean

the leading coefficient isn't a property dependent on any input

the leading coefficient is a property of this polynomial as a whole and is entirely unrelated to what values it may have at which points

[viscid thistle]

i dont think it depends on the input

[velvet blade]

Okay, it's just I thought that the properities get effected by some inputs like these

[viscid thistle]

because then f(x)=x^4+x^2 at f(1)=1^4+1^2=1+1=2=2^1 it would outright break

[sick steppe]

I wouldnt call a co-efficient a property of something, i'd call it a characteristic

[willow bear]

wow look at you spelling the word "coefficient" all fancy like with your fucking hyphen

why don't you throw a fucking DIAERESIS in there too

because OBVIOUSLY coëfficient is the Only Correcte Way™️ to spell the word

[sick steppe]

Literally do not give a shit ann

[willow bear]

(end rant)

if you havent noticed i dont give a shit about your attempts to draw a line in the sand either

[sick steppe]

At least i dont need to get angry when trying to help 🙃

[astral mantle]

[viscid thistle]

what does the fucking opposite face me

mean

[astral mantle]

in what context

[viscid thistle]

here

[rigid beacon]

Its used to be passive aggressive

[viscid thistle]

ohh

i am sorry when i text i dont use emoji's so i asked

i think i should really get with the times

lol i am only 16 and here i am talking like a boomer

[ripe adder]

I need help with a systems of equations problem

#76. I have so far figured out that:

(w+2)+( l-3)= -6^2

And

(w+1)+(l+2)=30^2

but when I try to solve its no solution

[willow bear]

yeah cause neither of your equations make any sense

why are you adding the width+2 and the length-3? why should this equal -36?

similarly, why are you adding the width+1 and the length+2? and why would that equal 900?

think about how you would find the area of a rectangle.

instead of just randomly mashing symbols together like you did there.

[ripe adder]

can you give me a hint

[willow bear]

think about how you would find the area of a rectangle.

[ripe adder]

as in

how do I apply that to this

[willow bear]

and also read the problem carefully. you need to extract more from it than just the numbers.

but first, tell me: if you denote with w and l the original width and length of the room, what is its area?

@ripe adder?

[ripe adder]

so it should be X and Y for length because thats what they use in the problem

im gonna swap to those

also

[willow bear]

lowercase x and lowercase y

[ripe adder]

it would be

[willow bear]

okay

x and y it is then

[ripe adder]

(x+2)*(y-3) = (xy)+30

(xy) = A

[willow bear]

see now we are getting somewhere

[ripe adder]

and the other one

[willow bear]

except you are mixing up the two plans

[ripe adder]

yeah

it shouild be

(x+2)*(y-3) = (xy)-6

and

(x+1)*(y+2)=(xy)+30

[willow bear]

there we go!

see when you give the problem some thought instead of just randomly mashing things together you get shit that makes sense

you should try it more often

so now you have the equations (x+2)(y+3) = xy - 6 and (x+1)(y+2) = xy + 30, are you able to solve these for x and y?

[ripe adder]

I used mathway to check if they are right

I could do substitution

[willow bear]

oh 😒

mathway huh

[ripe adder]

I dont cheat with it I swear

I just check to see if my progress is pointing towards the solution

[willow bear]

also i expected a "yes i am", "no i'm not" or "i have some ideas but am not 100% sure" or something

[ripe adder]

well I can do substitution method to solve for those right?

[willow bear]

you can always try

if it so happens that you can isolate one of the variables in one of the equations then there is no higher force stopping you from substituting it everywhere else

[ripe adder]

2x=-y+28

which means that

x=(-y+28)/2

Ok I solved it

thanks @willow bear

[glossy harbor]

I'm a bit rusty with inequalities. How can I show $n \geq 140 \implies \frac{n}{n-70} \leq 2$?

[obsidian monolith]

naveeex:

[glossy harbor]

Just by transforming the left-hand side, not by looking at the graph or intuition/substitution

For some reason I only come up with $\frac{n}{n-70} \leq \frac{n}{70}$

[obsidian monolith]

naveeex:

[earnest jungle]

$n-70$ is never negative in this case, so we have $n \leq 2n - 140$, or $140 \leq n$. Thus, this is true.

[obsidian monolith]

LifeSource:

[jagged orbit]

how would you prove that for x > 0, arccotx = arctan(1/x) ?

[earnest jungle]

$x = \cot\left(\tan^{-1}\left(\frac{1}{x}\right)\right) \Rightarrow x = \frac{1}{\tan\left(\tan^{-1}\left(\frac{1}{x}\right)\right)}$.

[obsidian monolith]

LifeSource:

[earnest jungle]

As you can see, from this we get $x = \frac{1}{\frac{1}{x}}$, or $x = x$. So there you go @jagged orbit

[obsidian monolith]

LifeSource:

[jagged orbit]

@earnest jungle What about something like this?

[earnest jungle]

That works @jagged orbit

[jagged orbit]

But then wouldn't you need to restrict x > 0?

Graphing arccotx and arctan(1/x) separately they're not equal when x < 0 @earnest jungle

But I'm not quite sure how to state that formally without saying I used something like graphing software

[earnest jungle]

I am not sure if there is a good way... Other than to say that they don’t intersect on $x < 0$

[obsidian monolith]

LifeSource:

[jagged orbit]

Then I guess that would have to be proven as well

I'm not sure how though

[earnest jungle]

I guess you can do it by contradiction and state that $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$ for $x \leq 0$, and show that this is not true

[obsidian monolith]

LifeSource:

[viscid thistle]

how do I do this,I tried using the triangle inequality but it didn't work, maybe there's smth I'm missing idk

[viscid thistle]

<@&286206848099549185>

[viscid thistle]

<@&286206848099549185>

[harsh smelt]

!r4

i mean ping only once

[viscid thistle]

what if the question has been unanswered for like 1 1/2 hours

[harsh smelt]

@viscid thistle

[obsidian monolith]

Commander Vimes:

[viscid thistle]

hmm

but how do we get the $<\varepsilon$ part

[obsidian monolith]

Alexander Grothendieck:

[viscid thistle]

idk how we get it from this

[harsh smelt]

i mean

[leaden stratus]

Hi. These exercises are about the undefined forms.

Number 134 isn't an undefined form and number 135 I don't know how to solve it

[sick steppe]

@leaden stratus both of them ask about end behaviour of polynomials

[earnest jungle]

That isn’t precalculus lol

@leaden stratus anyhow, for 135 the numerator (when plugging in $-\infty$) explodes to $-\infty$, right?

[obsidian monolith]

LifeSource:

[earnest jungle]

Dividing by two won’t change this

Thus that limit goes to $- \infty$

[obsidian monolith]

LifeSource:

[viscid thistle]

hello i have a doubt in l'hospital rule that why is it only valid in 0/0 or infinite/infinite limits and not in normal limits?
i have seen the proof according to that it shd be universal.....

[willow bear]

can you show the proof that you saw

applying l'hop to a limit that isn't in the form 0/0 or ∞/∞ can result in bullshit and it's very easily demonstrated

[viscid thistle]

uhm sorry its a little bit unclear as its my pc cam

[willow bear]

yeah? this will not work if the limit isn't of the form 0/0.

[viscid thistle]

why?

[willow bear]

you will not be able to subtract f(a) freely, for one

[viscid thistle]

if we see the proof then it shd be applicable in normal limits also right?

[willow bear]

the proof uses f(a) = g(a) = 0 as an assumption

[viscid thistle]

yeah i dont understand the thing why do they subract f(a)

means how do they put it 0

[willow bear]

they consider the 0/0 case plus a few extra assumptions

look do you want me to just give you an example of the bullshit that can result from applying l'hop blindly to a limit that isn't 0/0 or ∞/∞?

[viscid thistle]

but how is 0/0 related to equate f(a) and g(a)=0

[willow bear]

0/0 is shorthand for "the ratio of two functions both of which approach 0"

and we're assuming f and g are differentiable, hence they're continuous, hence they approach f(a) and g(a) respectively as x->a

[viscid thistle]

uhm wait

let me try to understand

[willow bear]

look do you want me to just give you an example of the bullshit that can result from applying l'hop blindly to a limit that isn't 0/0 or ∞/∞?

i think this will be the most productive way to move forward with this

[visual finch]

Is this solvable? https://i.imgur.com/smMqOnE.png

[willow bear]

what am i looking at and what do you want to do with this monstrosity of an expression

[velvet blade]

Hey I resumed my studies today, thankfully the lorries stopped dumping soil. I have trouble with 4th question

I don't understand how to solve it

[willow bear]

write $z = re^{i\theta}$

[obsidian monolith]

Ann:

[velvet blade]

yes I did

[willow bear]

one of the possible values for $w$ is $r^{1/n} e^{i\theta/n}$

[obsidian monolith]

Ann:

[velvet blade]

ok?

[willow bear]

call this w_0 for convenience

[velvet blade]

Okay

[willow bear]

notice that $w_0^n = z$ by construction, but also $(w_0 e^{2\pi i/n})^n = z$ too

[obsidian monolith]

Ann:

[velvet blade]

Okay

[willow bear]

this is not the whole answer ofc

i'm just trying to give you a push

[velvet blade]

Okay I'll try and let you know, thanks.

Wait that means rest of all the other complex numbers are inside the multiples of 0 and 2pi right?

Sorry

Let me rephrase it

Rest of the complex numbers have polar coordinates which have angles in the range of 0 and 2pi

[willow bear]

i didn't really say anything about theta. you can have it be between 0 and 2pi if you want but you don't need to.

the $n$ numbers you're looking for are given by $$w_k = w_0 e^{2\pi ki/n}, \quad k = 0, 1, \dots, n-1$$

[obsidian monolith]

Ann:

[velvet blade]

why have 2pi when it's equal to 1 though?

[willow bear]

since when is 6.28ish equal to 1

also note i am not involving e^(2πi) anywhere. maybe consider reading the whole thing i send instead of cherrypicking symbols

[velvet blade]

Ok I will. Give me 5

[velvet blade]

Ann I couldn't understand why to have a 2pi as an exponent, and how the numbers are said to be distinct when we're just multiplying one of the roots with a said exponent. Aren't we using one number to make another?

@willow bear

[willow bear]

Aren't we using one number to make another?
distinct doesn't mean unrelated

[velvet blade]

Okay but it'll just form one number right?

wait

sorry

I got it

What about 2pi as an exponent though?

I didn't get that part

[willow bear]

the exponent isn't 2π it's 2πki/n

or 2πi/n for my first thing

$e^{2\pi i/n}$ is a number that when raised to the $n$'th power gives 1

[obsidian monolith]

Ann:

[willow bear]

i.e. $e^{2\pi i/n}$ is an $n$'th root of unity.

[obsidian monolith]

Ann:

[velvet blade]

Okay got it

This was hard

ngl

[leaden stratus]

Yes, but I'd need the passages

[willow bear]

sounds like you overthought things a lot again @velvet blade

[velvet blade]

All of this feels new AF. I am unable to figure out how to think about a problem most of the time

[willow bear]

does the book teach you how to think of complex number multiplication as scaling & rotation

[velvet blade]

Polar coordinates?

5th question, is the answer $e^{ni\theta} = e^{2\pi i}$?

[obsidian monolith]

lazypawtato:

[willow bear]

no that is not the answer as written

[velvet blade]

Is it $e^{i\theta}= e^{\frac{2\pi i}{n}}$ ?

[obsidian monolith]

lazypawtato:

[willow bear]

hurgh

no

you're not doing what they ask of you

[velvet blade]

[willow bear]

what stopped you from at least writing out
1, e^(2πi/n), e^(4πi/n), e^(6πi/n), ..., e^(2(n-1)πi/n)?

[velvet blade]

I really don't know

wait why n - 1 in exponent? Isn't it supposed to be n in the end?

no wait

I can't

[willow bear]

if you go up to n you'll get e^(2πi), which is 1 and so you're taken right back to the beginning of the list

[velvet blade]

okay

I didn't solve any of these

[viscid thistle]

Im super confused by question c here

I dont see why they are using velocity x acceleration for these intervals

Acceleration should be the only thing that matters here

[leaden stratus]

What should I do here? Take out x^3 and have [x^3(9+X)/2]???

[sour hemlock]

I think the limit would be inf

[leaden stratus]

It is

But I NEED STEPS

[sour hemlock]

hmmm

well

[astral mountain]

just plug in - inf and see what happens

[leaden stratus]

I tried to do this

[sour hemlock]

just put inf in there

[leaden stratus]

If I plug -inf we have the undetermined form inf - inf

[sour hemlock]

I don't know what other steps

you can take the x^2 as common factor in numerator

[leaden stratus]

But isn't x^3 greater and dominate over x^2?

[astral mountain]

yes

[leaden stratus]

Then I would have to take out x^3 🤔

[astral mountain]

no

x^2(9x+1)/2

plug in -inf and see what happens

[leaden stratus]

Ok, I'm trying

[earnest jungle]

@leaden stratus I have already explained it man

[leaden stratus]

Yes, but I didn't understand it

[earnest jungle]

So basically, you focus on the higher powers in the denominator and numerator

The highest power in the numerator is x^3, right?

[leaden stratus]

Yes

[earnest jungle]

But we are going toward -infinity and the power is odd, so the numerator will explode to -infinity

[astral mountain]

@leaden stratus what u did in that pic is alrdy correct btw

[earnest jungle]

Since x^2, although will always remain positive, increases much slower

[leaden stratus]

@astral mountain yes, but we would have -inf * 9+0

[astral mountain]

so

thats just -inf

[leaden stratus]

-inf * 0 is undetermined

[astral mountain]

-inf (9+0)

thats -inf*9

[earnest jungle]

basically, just “plug in” -infinity to see what happens

[leaden stratus]

Ahhh

Omg

Yeah, it's 9+0

Due to stress I read 9*0

[earnest jungle]

From that, you extrapolate that the result is -infinity

Of the entire limit

[leaden stratus]

Hahaha thanks guys, I'm dumb 😅

[earnest jungle]

Np

[leaden stratus]

And what can I do here? I tried rationalising.

@earnest jungle

@astral mountain

[viscid thistle]

third step, it should be -x^2 not -x

remember that (a-b)(a+b)=a^2-b^2

[astral mountain]

I'd just pull out the x

[leaden stratus]

Ah right

At the denominator it's a mess

I'll probably have to pull +inf on every x

[viscid thistle]

if you fix it, you should get x^2-1-x^2 at the num

[leaden stratus]

Yes, fixed that

[viscid thistle]

which simplifies to something nicer...

[leaden stratus]

Yes, I already fixed that, thanks

[velvet blade]

Is induction just used to prove stuff? not derive it?

he just gave the equation for it. No derivation

[willow bear]

induction per se is not used for derivations, but the ideas behind induction naturally give rise to the idea of recurrence relations

and a recurrence relation may be analyzed (in a context-dependent manner) and solved

and the solution, if necessary, verified by induction

[velvet blade]

So binomial theorem isn't derived in highschool?

okayy

when is it derived though?

[willow bear]

there is a slightly informal combinatorial explanation for why the binomial theorem is what it is
is that what you're looking for?

depending on the highschool and exactly what goes into its math classes, the combinatorics may be accessible to its students

[velvet blade]

Okayy

So it's derived in combinatorics

[willow bear]

if you insist

idk like

there is a difference between arriving at a result and proving it

the former tends to be more informal and exploratory

[velvet blade]

Okayy

I assumed assertion is true and added n+1 numbers. Which gave me $1+3+5+...........+(2n-1)+2n = n^2 + 2n$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

I don't know how that says that it's true

$n(n+2)$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

Is that it?

[willow bear]

your last term isnt correct

also why not make a ton more dots in that ellipsis if you like it so much

1 + 3 + 5 + .............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. + (2n-1)

anyway

[velvet blade]

I didn't even make comically many

[willow bear]

the terms go up by 2 here not by 1.

[velvet blade]

Okay

[willow bear]

the (n+1)st term would be 2n+1

and you would have n^2 + 2n + 1, or (n+1)^2 as intended

[velvet blade]

I didn't understand