precalculus

[slow roost]

yessir

[fluid shore]

Good

[slow roost]

ok tysm!

i have a quiz tmorrow

so tyyy

[odd helm]

Mind if I ask now?

[slow roost]

lol sorry!

go ahead

[fluid shore]

Lel good luck for it

Don't memorize things, just think logically and you'll be fine.

[odd helm]

I'm not sure why it's 265

Shouldn't it be 180+4.57

Not 270-4.57

[wise kelp]

Why do you think it should be 180+4.57

[odd helm]

Because the reference angle is 4.57 and it was supposed to go in the southwest direction.

I can show you the beginning of the problem if you need it

[wise kelp]

If you're measuring the angle as shown in that image then it's 270-4.57

[odd helm]

But why though

You mean that arrow

[wise kelp]

Yeah

[odd helm]

Idk why she did that

Here I'll post the rest of the question

Hm actually maybe since it is going west that we go in the negative direction on the unit circle

Since if you go positive you are heading east

[wise kelp]

To my understanding, bearing is usually calculated clockwise from north

As in the first image you posted

[odd helm]

Oh

Well that explains it

Thank you

[wise kelp]

Np

[trim fable]

anyone here

[stable pendant]

No

We’re all dead

[trim fable]

lol

oof

[stable pendant]

Removed

[trim fable]

;-;

hey so

how would u do

15 b

hello?

[viscid thistle]

can you see what you are trying to solve for?

o this is late but @trim fable

[coarse wadi]

need some help with this 🙂

[viscid thistle]

in what universe is this precalculus

[coarse wadi]

idk too much drama in calc

[odd helm]

uh

[willow bear]

you were the one to start it fam

[odd helm]

Could I post a question

[willow bear]

yes

flub's doesn't belong here

[odd helm]

Sorry

[languid crane]

Why is this always full of calc questions

It’s called PRECALC for a reason

[willow bear]

people see calculus and forget about the pre

[harsh cipher]

For this question...

I use doubling time formula.

first step is to find how many times bigger it would be compared to 25 days doubling time by substituing 60.

[willow bear]

I use doubling time formula.
don't

[harsh cipher]

then divide by doubling time which is P= 2PO

[willow bear]

you can immediately write down the formula for the population at t days knowing that the doubling time is 25 days.

$P(t) = P_0 \cdot 2^{\frac{t}{25}}$

[obsidian monolith]

Ann:

[harsh cipher]

Ann, you are next level thinker. 😛

[willow bear]

i suppose you could call it next-level if you mean it to be above the level of "blindly applying formulae with little to no intuitive understanding of what's going on"

[harsh cipher]

I still don't get it

haha

[willow bear]

ok so like

given that formula i posted there

[harsh cipher]

I need through explanation.

[willow bear]

P(t) is the population after t days

and i claim it's given by that formula

i invite you to calculate P(25) using it

and see that it works out to 2P_0

exactly as it should

report back when you do so, and tell me whether or not it makes sense to you.

[harsh cipher]

Nope I didn't get it.

[willow bear]

...

ok, so what did you get for P(25), if not 2P_0?

[harsh cipher]

4.723

[willow bear]

wh

no

how

how did you even get that

[harsh cipher]

oh

2.36

[willow bear]

what

[harsh cipher]

that's what I got

[willow bear]

no!!!

NO!!!!!!

$P(t) = P_0 \cdot 2^{\frac{t}{25}} \ P(25) = ; ?$

[obsidian monolith]

Ann:

[harsh cipher]

I'm laughing too hard right now, I feel terrible

[willow bear]

answer my question, please

[harsh cipher]

p(25)= P_O x 5.27

[willow bear]

...

so you're saying $2^{\frac{25}{25}} = 5.27$???

[obsidian monolith]

Ann:

[harsh cipher]

no

[willow bear]

then how are you getting that 5.27 figure

[harsh cipher]

i subbed 60 and lolol

[willow bear]

what

why did you sub 60

[harsh cipher]

sorry Im not focusing rn

lemme do it again

[willow bear]

i told you

exactly what to do

[harsh cipher]

P(25) = P_O x 2

[willow bear]

P_0, not P_O.

but anyway

yes

$P(25) = 2P_0$

[obsidian monolith]

Ann:

[harsh cipher]

okay

[willow bear]

which is exactly what you would expect given that 25 days is the doubling time

does this make sense to you

i am not going to continue until i get an indication from you that it makes absolute perfect 100% sense

[harsh cipher]

so after we divide that by

[willow bear]

i am not going to continue until i get an indication from you that it makes absolute perfect 100% sense(edited)

[harsh cipher]

yes

the above part makes sense

[willow bear]

ok, great.

so do you understand why the population can be modelled by $P(t) = P_0 \cdot 2^{\frac{t}{25}}$?

[obsidian monolith]

Ann:

[willow bear]

i.e. that that equation does indeed describe exponential growth as i claim it does?

[harsh cipher]

Yes it does

only after I learned that you need to put 25 inside of P(t)

sigh...

[willow bear]

no you don't

[harsh cipher]

why?

[willow bear]

you don't NEED to do anything

i instructed you to calculate P(25) as a means of illustrating it to you

[harsh cipher]

that's what I'm saying

[willow bear]

ok

so

well at this point

the answer to your original problem is $\frac{P(60)}{P(25)}$

[obsidian monolith]

Ann:

[harsh cipher]

Yea

[willow bear]

and i hope it makes absolute 100% perfect intuitive sense to you WHY that expression gives you the answer

whatever its value may be

[harsh cipher]

So what I do is after finding the population after 60 days.

Then divide by we divide by what you first taught me.

Then divide by what you first taught me

https://cdn.discordapp.com/attachments/363224154469826562/654973392205905920/183668144404037632.png

that

🙂

[willow bear]

well

i suppose that's a way to word it

[harsh cipher]

hahahaha

how would you word it

[willow bear]

i already did

the answer to your original problem is P(60)/P(25)

you said "so what i do is after finding P(60) i divide it by P(25)"

[harsh cipher]

yea

[willow bear]

which i guess is just a rephrasing of what i said, but in words

and way less compact

[harsh cipher]

Thanks Ann

really appreciate your time for helping me understand.

[willow bear]

you're welcome

i hope this sticks

[zealous sail]

I don't understand that step

[fluid shore]

They factored out 1/(x-a) and put everything inside over a common denominator

Work through it on a piece of paper lmao

[willow bear]

$\frac{1}{x^2(x-a)} - \frac{1}{a^2(x-a)} = \frac{1}{x-a} \paren{\frac{1}{x^2} - \frac{1}{a^2}}$

[obsidian monolith]

Ann:

[willow bear]

do you understand this @zealous sail

[fluid shore]

It might be more fruitful to allow him to work through it on paper first before giving him that step.

But sure

[zealous sail]

I've never factored out a whole fraction before

so this is new

I am going through it on paper now

[fluid shore]

Yeap

With these sorts of things, take your time. There’s no need to rush.

[zealous sail]

Am i even on the right track?

this looks worse than what I started with

[fluid shore]

Well you certainly made it more complicated

Okay, come write your problem out first on a piece of paper and send it here

[zealous sail]

Well, I tired having the same denom. for both first

[fluid shore]

Well, that’s certainly fine but it’s certainly more complicated

[zealous sail]

This is the problem. make the left side look like the right

[fluid shore]

Okay, now, do you see the fact that 1/(x-a) is a common factor for both fractions?

[zealous sail]

Yeah, what Ann said, is now obvious to me

[fluid shore]

Yes

So pull them out

[zealous sail]

I got it now

[fluid shore]

Ann made it rather obvious but you have to be able to see these kinds of things on your own, yea?

[zealous sail]

My problem was not factoring part of the fraction out of itself, I only looked to factor stuff in the num and denom separately.

Like I said, never have I ever had to do that

[fluid shore]

It’s okay

[zealous sail]

thanks @willow bear @fluid shore

[ripe dust]

Can someone tell me where I messed up

[willow bear]

the very first step

[ripe dust]

R.I.P.

[willow bear]

you can't go from a = b/c to a/c = b

[ripe dust]

Thanks ooof

geez I always do dumb algebra mistakes in calculus

[rich ocean]

@ripe dust what you probably intended to do was:
a = b/c => 1/a = c/b => b/a = c
(where a = 1.3, b = 5, c = 2 + 8e^(-0.75t))

[ripe dust]

Hmmm looks interesting

Yeah

[nova dew]

ok so I'm trying to derive $\frac{d(x^n)}{dx}=nx^{n-1}$ for the first time without looking it up

[obsidian monolith]

CoolShot:

[nova dew]

I did some stuff but didn't exactly get it

can someone tell me where I went wrong

here's what I did

[viscid thistle]

Oh that’s easy

So simple lmao

Tf

[nova dew]

f(x)=x^n, I took a point (x+h, f(x+h)), then I did $\lim_{h\to0} \frac{(x+h)^n -x^n}{x+h - x}$ then I used binomial expansion and $\lim_{h\to0} \frac{x^n + ^{n}C_1 x^{n-1}h + ^{n}C_2 x^{n-2} h^2 + \cdots + h^n - x^n}{h}$ and then one h cancels out in each to give $\lim_{h\to0} x^{n-1} + ^{n}C_2 x^{n-2}h^2 + \cdots$ and then all the terms with h will be set to zero so I end up with $x^{n-1}$

where am I missing an n

did latex bot break where did it go

[wise kelp]

@viscid thistle Can you like, not do that

Just because something's simple for you doesn't mean it's simple for everyone else

[nova dew]

ok since the bot decided to die I'll just post the image of my working

[pale kettle]

Check that you applied the binomial formula correctly

[nova dew]

oh

well that was a stupid error but oh well

thanks

Next I'm going to try d(sin(x))/dx=cos(x) which is probably a few miles harder but oh well I'll try it out

[valid violet]

You'll need sin(x)/x as x to 0

@nova dew

[nova dew]

nice I give up now

no wait I have a good idea I'll try it this time

oh wow I actually did it

am I mathematician yet

$\lim_{h\to0}sec(x+h) = sec(x)$ is this correct

[obsidian monolith]

CoolShot:

[odd helm]

Why did they find a50 shouldn't it be a60?

[viscid thistle]

i think this makes no sense in multiple ways

its like all 50s and 60s were erased and a random choice of 50 or 60 was put in place of it

what is this problem and solution from @odd helm

[serene heath]

sum of first 60 terms of a series

[hearty spire]

Hello, how to write this complex number in algebraic form?

[willow bear]

$i = e^{i\pi/2}$, so your thing would be $e^{-3\pi/2}$ i suppose

[obsidian monolith]

Ann:

[hearty spire]

thanks

[odd helm]

I’m lost why did they raise it to the 1000 power? Isn’t this sequence going forever since the triangles will keep being divided by 4

[viscid thistle]

have you learned about converging and diverging series yet

raising it to the 1000th power is uh... not doing anything but trying to make an intuitive/inductive reasoning point on why it converges

the question is kind of poorly phrased

you cant quantify “all” the black triangles

well actually i guess thats fine to say

but without concepts of limits it may not make total mathematical sense why the answer to a sum of infinite black triangles is exactly 1/3

[native timber]

you can do it algebraically

$S = \frac{1}{4} + \frac{1}{4}\cdot S$

[obsidian monolith]

Jiramide:

[ripe dust]

baby mode algeb

[native timber]

$4S = 1 + S \rightarrow 3S = 1$

[obsidian monolith]

Jiramide:

[viscid thistle]

well yes you can do it by throwing it into the geometric series formula but thats not really going to help in understanding

[native timber]

fair enough

[viscid thistle]

https://i.imgur.com/3itVHpk.png this kind of demonstrates how you can add infinitely many things and it be equal to a finite number

no matter how many finitely many terms you add you never reach 2

probably easier to see than trying to imagine the triangle example

[native timber]

wait, is he asking why they used S_1000 as the "final element" (using sum of geometric sequence formula) when S is an infinite sequence?

[viscid thistle]

yea that was part of it at least

[slow roost]

For the 22) problems

How could I check my answers on a calculator

[serene heath]

,rotate

[slow roost]

So I plugged those in a calculator as

Tan^-1(-1/.658)

And then went on to get mu two angles which are

Radian*

7.272 and 2.152

How do I check if that is equivalent to the original values

Using a calculator

[serene heath]

just find the cot of that angle

both of them

[slow roost]

Let me see

[serene heath]

and if they give you -0.658 ur good to go

[slow roost]

Hold up

[viscid thistle]

i dont think your answers are gonna work out

[slow roost]

My answer key says those are the answers tho

@serene heath that’s what I did but it doesn’t match

[viscid thistle]

its because

1. the answer key is wrong

[serene heath]

u need to use the exact values

oh rip

[viscid thistle]

2. your -1/ is inside the function

[slow roost]

So what would the answer be then

5. Something?

[serene heath]

,w solve cotx=-0.658

[obsidian monolith]

Display results online and refine query

Requested by the one n only#3070

[slow roost]

That’s what my key says

[viscid thistle]

oh thats actually right

my bad

[slow roost]

Mhm

[serene heath]

7.2 isnt even in the solution range lol

second one is fine tho

[slow roost]

But how would I check it on a calculator

Shouldn’t it have bee

5.727

I initially got

5.294

[serene heath]

yea

5.294 is right

[slow roost]

Mhm

Is the second one incorrect

Cause-

[viscid thistle]

2.152 and 5.294 work

[slow roost]

Ok

How would I check in my calculator

Lol cause I’m doing something wrong

Oops

Cause that’s what I get back

[viscid thistle]

just do 1/tan(2.152)

thats the same as cotangent(2.152)

[slow roost]

Oh

Thank you!!

[odd helm]

I have a question does it always diverge if it is arithmetic

I mean since we are constantly subtracting it is heading for -infinity right?

So there can’t be a sum

[warped willow]

then write diverges?

[dim jungle]

https://www.youtube.com/watch?v=w-I6XTVZXww

[viscid thistle]

bruh

[wise kelp]

It's true, I calculated it

[frozen needle]

[valid violet]

what's true

[wise kelp]

That 1+2+3+4+5...=-1/12

It turns out that if you add up all the numbers, you do in fact get -1/12

You just gotta be persistent

[ripe dust]

That's dedication there

keep going up

[zealous sail]

shouldn't zero be included on the bottom one?

or really there should be another line 0 x = 0

[native sequoia]

we like to keep the sub-domains disjoint

[zealous sail]

what is that in simple english?

[native sequoia]

idk
by subdomains I'm referring to x>0 etc..
and the pic is correct

[pale kettle]

It doesn't matter

The function is defined at x = 0

Since it is part of the first part

There's no reason to define it again in the second part

[viscid thistle]

For question 8, why do I need to mutually by ( x - 3)squared

I mean if I just mutually by (x -3 ) I only get one answer?

Multiply *

[uncut mulch]

there are multiple ways to approach it
it is multiplied by (x-3)^2 here since it is known to be positive

[viscid thistle]

?

[uncut mulch]

and won't affect the direction of the inequality

[viscid thistle]

Why can’t I just multiply by (x-3) then?

[patent beacon]

Ooh that's clever

I've never seen it approached that way

[viscid thistle]

Bruh

[uncut mulch]

because x-3 isn't positive for all x (x != 3)

[viscid thistle]

Yeah?

[patent beacon]

So if you multiply or divide by a negative quantity, you have to switch the inequality

[viscid thistle]

Oh is is because x - 3 will be always be positive when squared? That’s why we multiply by that?

Ok

[patent beacon]

You can multiply by x - 3, but you have to do two cases for the different possibilities

[viscid thistle]

I see thx

[uncut mulch]

turning it into a quadratic inequality allows you to solve it graphically

[viscid thistle]

Oh but if I multiply by (x-3) how do I do the two cases?

[patent beacon]

Do one where x - 3 is positive
And another where x - 3 is negative

[uncut mulch]

or factorise/ rough sketch

what do you get when you multiply it?

[viscid thistle]

You can’t just flip the sign tho

Question 8 answer

[uncut mulch]

yeh. but try getting there yourself, i'm guiding you through this method

step1: multiply both sides by (x-3)^2

[uncut mulch]

no, that isn't multiplying both sides by (x-3)^2

(that is multiplying both sides by x-3, assuming x-3 is positive)

[uncut mulch]

be careful with the direction of the inquality

[viscid thistle]

Yeah

[uncut mulch]

after subtracting both sides by 5(x-3), you'll get **stuff > 0 **

[viscid thistle]

Oh

[uncut mulch]

also it was unecessary to expand like that

[uncut mulch]

it is easily factoriseable in the form
2**(x-3)^2 - 5(x-3)** > 0

[viscid thistle]

Oh

I see

Thx anyway

[uncut mulch]

alternatively if you want to go down the route for just multiplying by x-3, then
FOR x-3 > 0 → x > 3
5 < 2x - 6 → x > 11/2
x > 3 AND x > 11/2 → x > 11/2
FOR x-3 < 0 → x < 3
5 > 2x - 6
x < 11/2
x < 3 AND x < 11/2 → x < 3

[viscid thistle]

Oh

Cool

[uncut mulch]

another method is to combine them into a single fraction and analyse signs

$\frac {2x - 11}{x-3} > 0$

[obsidian monolith]

ramonov:

[viscid thistle]

I actually learnt in equality last year and I was doing practice question and stumbled Upon this one

The textbooks are not in depth enough tho

[uncut mulch]

just be careful when multiplying inequalities with unknowns

[slow roost]

can someone help me with this

That’s all I have right now

And no, I don’t know what the addition formula for sine is

[valid violet]

so go learn it

[slow roost]

...

lol

my test is tomorrow and my class is the only class that was given this lesson which did not go over much

[odd helm]

https://m.youtube.com/watch?v=sU2pyMR8GZ4

You can use this to learn it ig

[slow roost]

i learned that

[odd helm]

Oh

[slow roost]

except for the tangent one

only did sin and cos

[odd helm]

Oh well the tan is in that vid so if you need to learn that you can use this vid

[little owl]

I never used/learned the tangent one

[odd helm]

I did

But I forgot it since I learned it a while ago

Sin and cos are easy to remember

[slow roost]

yeah

Cos is just opposite of what u would think

but idk how to apply it to 15

would i convert everything to angles

and then ask what combinations make up sin pi/2 aka 90º?

[odd helm]

I mean the sin of pi/2 = 1 and the cos= 0

[slow roost]

uhuh

[odd helm]

So (1)costheta +(0)sintheta

Right?

[slow roost]

but its sin pi/2 + sintheta= cos theta

sin pi/2 is 1

so isnt that 1+sintheta=cos theta

[odd helm]

I thought the formula was sinxcosy +sinycosx

[slow roost]

yeah it is

hm i see what ur doing

[odd helm]

So if it is sin(90+ theta) it would be sin(90)cos(theta)+cos(90)sin(theta)

[slow roost]

mhm

which is equivalent to

sin1costheta + sinthetacos1

right?

[odd helm]

It would be sin(theta)(0) because cos90=0

But the other part is correct

[slow roost]

oh ok

[odd helm]

So 1 times cos(theta) just leaves cos(theta)

[slow roost]

ok

thank you!

that makes sense

[odd helm]

You’re welcome

[slow roost]

Can I get help with 16

[untold cliff]

Let cos(theta) = u and solve for u. As u = cos(theta) it will equal the solutions to u, should be 2 of them, find theta

[slow roost]

there 4 of them

[untold cliff]

2u^2=1

square root, so +, - answer

u=sqrt(1/2), u = -sqrt(1/2)

[slow roost]

so would it be

cosroot2/2

so i got

pi/4 and 7pi/4

but then u have to realize isnt it cos= root2/2 and cos -root2/2

so u get 4 answers

[untold cliff]

oh i thought u were talking about before that, ya

[slow roost]

so the answers are

pi/4,7pi/4,5pi/4, and3pi/4

[untold cliff]

seems fine

[slow roost]

Is #6 4200pi?

[viscid thistle]

Find cos<E

Is there enough info to solve this?

@viscid thistle what letter is that in the bottom left

D

@viscid thistle

Ahaha, either way, no, I do not believe that is solveable without more information

are you sure you copied it down correctly?

Yup I drew the exact same thing as on my paper, the question literally only says "Find cos<E"

what does cos<E mean?

There's nothing even in the previous problem that relates to this one

Cos of the angle E

[warped willow]

yea no

[viscid thistle]

How to do this question? I set delta greater than 0 and didn’t know how to continue.

[viscid thistle]

[willow bear]

what did you get for delta

[viscid thistle]

This

@willow bear

dude

[willow bear]

alright well

you want this inequality to be true no matter what a is

[viscid thistle]

yes

[willow bear]

so

ITS discriminant must be negative

[viscid thistle]

?

what

why

[willow bear]

for the exact same reason that's true for any other quadratic

[viscid thistle]

doesn't this mean there is no maximum number?

ann

@willow bear

if the discriminant is negative, that means there are no real roots?

[willow bear]

hhhhhhhhhhhhhhhhhh

no look

[viscid thistle]

;;

I dont get why

[willow bear]

the discriminant being negative means that a^2 + 4a - 8b - 12 = 0 never happens

which is exactly what you want

[viscid thistle]

oooohhh

dang

if I got this question right

my concepts are that great yet I guess ...

[summer monolith]

Is this true?

[willow bear]

$a(bc) = bac$ yes

[obsidian monolith]

Ann:

[willow bear]

because multiplication is associative and commutative

[summer monolith]

Why lna remains unmultiplied by a^x^2 ?

[willow bear]

what

do you think a * (b * c) should be equal to (a * b) * (a * c)?

[summer monolith]

I think I confused (2x*lna) with (2x+lna)

sry about it

[willow bear]

yes you did.

[green zenith]

What do they mean by eliminating x from the 2nd equation by adding -2 times the first equation to the second?

[wise kelp]

,rotate

[wise kelp]

Ah, so what they're saying is

The first equation is $x-y=1$

[obsidian monolith]

GrandPiano:

[wise kelp]

You can multiply both sides by -2 to get $-2x+2y=-2$

[obsidian monolith]

GrandPiano:

[wise kelp]

And then adding this to the second equation gets you $2x+y+(-2x+2y)=6+(-2)$

[obsidian monolith]

GrandPiano:

[wise kelp]

Which simplifies to $3y=4$

[obsidian monolith]

GrandPiano:

[wise kelp]

@green zenith

The goal is to get an equation with just one variable, so if you make it so one equation has -2x and the other has 2x, you can add them together and the x's will cancel out, leaving you with just y

Though I don't know why they didn't just add the equations together as they are and eliminate y

[green zenith]

Ahhh now i get it

I wasnt familiar with adding equations

So i literally just add lh to lh and rh to rh when adding equations?

Anyway, thanks for the answer @wise kelp

[wise kelp]

Yeah, since the left hand side and the right hand side are equal you're still adding the same amount to both sides

Np

[blissful wadi]

any help with this limit please

i prooved this before but idk how to use it to get the limit

[serene heath]

what if f(x)=arctanx-x

[serene heath]

ok you can use that

find the limit of f(x)

then link it back to your limit

[blissful wadi]

any ideas about how to solve this one ?

$\frac{\arctan(\frac{x}{x+1})-x}{x^2}$

[obsidian monolith]

dvaix:

[blissful wadi]

x goes to 0

[frozen needle]

You can rewrite it as $\displaystyle\frac{\arctan!\prn{\frac x{x+1}}-\frac x{x+1}-\frac{x^2}{x+1}}{x^2}$

[obsidian monolith]

Tuong:

[frozen needle]

$\displaystyle\lim_{x\to 0}\frac{-\frac{x^2}{x+1}}{x^2}$ is easy to deal with, the difficulty lies in the calculation of
$$\lim_{x\to 0}\frac{\arctan!\prn{\frac x{x+1}}-\frac x{x+1}}{x^2}$$

[obsidian monolith]

Tuong:

[frozen needle]

$$\lim_{x\to 0}\frac{\arctan!\prn{\frac x{x+1}}-\frac x{x+1}}{x^2}=\lim_{x\to 0}\frac{\arctan!\prn{\frac x{x+1}}-\frac x{x+1}}{\frac{x^2}{(x+1)^2}}$$

[obsidian monolith]

Tuong:

[frozen needle]

I multiplied by something that tends to 1, it doesn't change anything

but then this limit can be seen as

$$\lim_{t\to 0}\frac{\arctan(t)-t}{t^2}$$

[obsidian monolith]

Tuong:

[frozen needle]

and that looks a lot cleaner

[obsidian monolith]

Whoever:

[frozen needle]

would you be able to calculate that limit, @blissful wadi ?

[viscid thistle]

what

[blissful wadi]

hmm

one sec

[viscid thistle]

isnt that what he started with

[blissful wadi]

yeah

isnt that what i started with ?

it will give me 0 times +inf

just like the first expression

doesnt it ?

[frozen needle]

it's a lot different

now it's easy to do manipulations on it

[blissful wadi]

by seperating the two terms

?

[frozen needle]

no

[blissful wadi]

hm

[frozen needle]

what do you know about arctan?

[blissful wadi]

reverse function of a restriction of tan

arctan (t) = -arctan (-t)

lim x goes to 0 arctan(x)/ x = 1

lim in 0 is 0

[frozen needle]

most of this stuff isn't very useful

[blissful wadi]

well thats all we know atm about arctan

i dont see the idea tbh

in your expression

for me its similiar to the original one

[frozen needle]

what do you know about arctan's regularity?

[valid violet]

If it helps, (arctan(t)-t-(arctan 0 -0))/(t-0) as t goes to 0 is the derivative of arctan -t at 0

[blissful wadi]

"arctan's regularity" dont know what thats is

[frozen needle]

continuity, differentiability, etc.

[blissful wadi]

ah

continious on IR

the derivative of it is 1/1+x^2

of arctan ( x)

[frozen needle]

So you can write $\arctan t$ as $\int_0^t\frac 1{1+u^2}\dd u$ for all $t\in\bbR$

[obsidian monolith]

Tuong:

[blissful wadi]

oh

we didnt study integrals yet

[frozen needle]

ach

there's a way around

[valid violet]

I like my idea ✋

[blissful wadi]

@valid violet the derivative of (arctan(t)-t)

?

u wanted to say

[frozen needle]

do you know Rolle's theorem?

[blissful wadi]

yes

[valid violet]

Yeah i fixed it as you were correcting me

[blissful wadi]

yeah that's a better ideaa i see it working

[valid violet]

But

If you're not allowed to use lhr

I'm not sure this is allowed

It's similar to lhr

[blissful wadi]

lhr ?

lhopital u mean ?

[valid violet]

Lhopitals

[blissful wadi]

if thats what u mean i've see it being used once or twice by my prof even tho he said there always a way arround but i guess its acceptable

@frozen needle what about rolle ?

[frozen needle]

you can prove a very cool theorem using Rolle

you can show that for all $t\in\bbR$, there exists $c_t$ strictly between $0$ and $t$ such that
$$\arctan(t)-t=\arctan'''(c_t)\frac{t^3}6$$

[obsidian monolith]

Tuong:

[blissful wadi]

whats Ct ?

[valid violet]

some constant

[frozen needle]

some number that exists

[blissful wadi]

aah

thought its some kind of a function

[frozen needle]

it depends on t so that's why I put a t in the index

[blissful wadi]

oh yeah

[frozen needle]

to prove that,

fix $t$, and have a look at the functions $\delta:\bbR\to\bbR$ and $\Delta:\bbR\to\bbR$ defined by
$$\forall u\in\bbR,\quad\delta(u)=\arctan(t)-\arctan(u) - \frac 1{1+u^2}(t-u)-\frac{2u}{(1+u^2)^2}\frac{(t-u)^2}2$$
and
$$\forall u\in\bbR, \quad\Delta(u)=\delta(u)t^3-\delta(0)(t-u)^3$$

[obsidian monolith]

Tuong:

[frozen needle]

Rolle's theorem applies to Δ

[blissful wadi]

oh so you used the theoreme of " acroissements finis " ( idk the name of it in english ) and made a new fuction of it

[frozen needle]

ah tu es français !

[blissful wadi]

marocain plutot mais on etudie en langue francaise

[frozen needle]

vous avez fait l'inégalité de Taylor-Lagrange ?

[blissful wadi]

non pas encore

[frozen needle]

Bon, grossomodo, le théorème de Rolle peut s'appliquer à Δ, car Δ(0)=Δ(t)=0 et Δ est dérivable partout

donc il existe c quelque part entre 0 et t tel que Δ'(c)=0

et quand tu calcules Δ'(c), ça te donne l' égalité annoncée avant

[blissful wadi]

mais ca cest coté theorique plutot pour deduire que la limite existe

nest ce pas ?

ma question est de calculer la limite

[frozen needle]

quand on a fait beaucoup de limites, on prédit que (arctan(t)-t)/t² ça tend vers 0 quand t tend vers 0

ou quand on connaît un peu plus sur arctan et les dérivées

[blissful wadi]

plus evidente puisque h(x) = 0

lorsque x = 0

plus une simple application de hopitale ocnfirme que cest 0

mais comment lacalculer cest ca le prob

[frozen needle]

T'as réussi à suivre tout depuis le début ou il faut recommencer ?

[blissful wadi]

franchement j'ai pas bien compris mais la ya la methode de montrer que c'est le derive de (arctan (t) - t ) en 0 et j'ai encore 4 question a faire

sinon ta pas a video qui explique cette methode que tu viens de faire ?

[frozen needle]

Le truc de la dérivée ça aurait marcher s'il n'y avait que t au dénominateur, mais là ya t²

Le théorème c'est "égalité de Taylor-Lagrange"

Tu peut aller voir la démo sur wikiversity

c'est juste le théorème de Rolle appliqué à une bonne fonction

[blissful wadi]

ah cest un theoreme j'ai cru que cest une petit conclusion de rolle de ta part xD

[frozen needle]

mais le théorème le plus pertinent ici serait quand même la formule de Taylor-Young

avec les développements limités tout ça tout ça

avec les développements limités, tu fais direct arctan(t)=t+O(t³) quand t tend vers 0 et pis voilà

[blissful wadi]

les developpements limités nous donne une fonction egale ou presque egale ?

je peux substituer arctan (t) par t+o(t^3) pour calculer la limite

[frozen needle]

Le développement limité donne une description du comportement asymptotique

c'est pas une égalité, il faut faire attention

et pour le o(t³), il faudrait écrire un terme -t³/3 en plus

terme qui est mangé si tu écris le développement avec O(t³)

[blissful wadi]

franchement ca me fait du mal a la tete , evidemment ton niveau est plus haut que le mien alors ces methode la sont normal pour calculer une limite pour toi

pourtant la pour nous c'est hort programme tous ca

et si je me rappelle bien on utilise ce developpement limité pour encadré la fonction orignal nest ce pas ?

[frozen needle]

on utilise un développement limité pour obtenir une description d'un comportement asymptotique

[blissful wadi]

oui je sais , le cas pour une asymptote oblique par exemple

mais la on l'utilise que pour tracer des courbes

[frozen needle]

ça sert pas trop à tracer des courbes les développements limités

x')

[blissful wadi]

bah a mon niveau cest que pour les Cf

ptdr meme la methode de derive ne marche pas a cause de t^2 j'ai pas fait attention au debut

j'ai fait 2h sur une limite 😂

[blissful wadi]

oh

managed to get it

@frozen needle Je me suis souvenu que le prof nous a donner cette inegualité pour nour aider dans un exercice similaire

[frozen needle]

ah oui c'est pas mal ça

[blissful wadi]

c'est ca pk j'ai dit avant que les developpelemnts limités nous donne un encadrement

[frozen needle]

$|\cos(t)-1|\leq\frac{t^2}2$ et $|\sin(t) -t|\leq\frac{|t|^3}6$ sont plutôt pas mal aussi

[obsidian monolith]

Tuong:

[blissful wadi]

noté meeerci

[frozen needle]

Le résultat balèze derrière tout ça c'est vraiment l'inégalité de Taylor-Lagrange

[blissful wadi]

c'est au sup ca

[frozen needle]

Oui

[blissful wadi]

chui enecore en bac SM

je sais pas c'est quoi lequivalacne de bac marocain au france

2eme annee bac plutot

[frozen needle]

moi non plus, ils ont changé beaucoup de chose depuis l'année où j'ai eu mon bac

(en 2017)

[blissful wadi]

votre bac c'est la derniere annee du lycee ?

[frozen needle]

oui

[blissful wadi]

a quelle age ?

[frozen needle]

les plupart des gens ont leur bac l'année de leur 18 ans

[blissful wadi]

ah

meme chose pour nous

17 / 18

https://www.alloschool.com/element/68480

exam national pour le bac marocain

[odd helm]

Why is the answer negative?

y is being raised to the 12 so

Wouldn't that make it a positive?

[wise kelp]

It’s not $\left(\left(-y\right)^4\right)^3$, it’s $\left(-y^4\right)^3$

[obsidian monolith]

GrandPiano:

[wise kelp]

$-y^4=-(y^4)$

[obsidian monolith]

GrandPiano:

[wise kelp]

@odd helm

[odd helm]

Oh

Thank you that makes sense

[ripe dust]

Huh does someone know where the proof for the identity is

I don't know how I could have solve without knowing that

also tried to search on google

Can't find anything abot it

or if you got another way to solve it just lmk

[fluid shore]

Search for the proof of the addition identities for trigonometric functions

You should get something for that

[ripe dust]

👍 ty

got it

geez so many stuff to remember for calc

does anyone know if there's another way to solve what I posted without using that proof?

[fluid shore]

Don’t forcefully memorize it. Just focus on learning it properly and it’ll automatically come to you

[ripe dust]

Ah god I tried to replace sinx by sqrt(1-cos^2x) but it ended up in some horrible newton raphson shit 😂

this identity is fairly easy to remember I just have to set y to x

[fluid shore]

Well you can derive it and you’ll end up remembering it

[harsh cipher]

Hi

do I make the bases the same?

[undone pawn]

yes

[harsh cipher]

I wans't sure if I had to use properties of logs extended power rule

[undone pawn]

bring it to the form $2^{\log_{2} x^a} $

[obsidian monolith]

FlynnXD:

[harsh cipher]

ok then does 2 log 2 get cancelled

and left with 3x^27?

😦

[undone pawn]

no it doesn't work like that

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

[obsidian monolith]

FlynnXD:

[undone pawn]

you're misusing this

you can't just bring the 3 along with it

get rid of the 3 somehow

[harsh cipher]

well.

when we make log base 8 into base 2

we get 2^3....

2^log base 2 get cancelled...

We're left with 3 x 27 ^ x?

I get confused where the 3 goes after we make the bases the same.

[stuck lark]

i see two ways to do this. rewrite the log in the exponent into base 2 or rewrite the base of the exponent to base 8

[undone pawn]

^

@harsh cipher do you know this $\log{2^a} = a\log{2}$

[harsh cipher]

thats power rule in reverse

[obsidian monolith]

FlynnXD:

[harsh cipher]

thats the same thing

[undone pawn]

use this to get rid of the 3 you have

right now you have $2^{\frac{1}{3}\log_{2} x^{27}}$

[harsh cipher]

how can u bring the 3 infront of log

that was the power of a base

[undone pawn]

whoops

[obsidian monolith]

FlynnXD:

[harsh cipher]

see that's where I'm getting confused

[undone pawn]

good catch

its 1/3

[harsh cipher]

okay..I still don't get or forgot how to bring the 1/3 infront of log

[undone pawn]

$\log_{b^n} a = \frac{1}{n}\log_{b} a$

[obsidian monolith]

FlynnXD:

[harsh cipher]

im looking at this

and im going crazy haha

https://www.symbolab.com/solver/factor-calculator/2^{\log_{8}\left(x\right)^{27}}

ok I need to figure out what above formulae was...

what is the name of that formula

[undone pawn]

let $\log_{b^n} a =c$

[harsh cipher]

log base b n a = 1/n log base b a

[obsidian monolith]

FlynnXD:

[undone pawn]

then by definition, $(b^n)^c = a$

[obsidian monolith]

FlynnXD:

[undone pawn]

so $b^{nc} = a$

[obsidian monolith]

FlynnXD:

[undone pawn]

take log base b on both sides now

so you'd get $\log_b b^{nc} = \log_b a$

[obsidian monolith]

FlynnXD:

[undone pawn]

so $nc = \log_b a$

[harsh cipher]

Im very very confused to tell the truth

[obsidian monolith]

FlynnXD:

[undone pawn]

so $c=\frac{1}{n}\log_b a$

[obsidian monolith]

FlynnXD:

[undone pawn]

hence proved the rule i just told you

[harsh cipher]

If this question is not on the test i will not be happy that's for sure.

either way I want to learn how to solve it

[undone pawn]

alright

[harsh cipher]

damn it it's taking up a lot of my time...sigh 😛

[undone pawn]

just clear your mind and focus

[harsh cipher]

k

[undone pawn]

see the rule i wrote above, and see the next few lines for a proof

use that rule, the 3 comes out

and then use what you call the "power rule" to simplify

[odd helm]

I used combinatorics for part 1 and permutation for part 2

Could someone explain why it’s in the opposite order?

[heady jewel]

What do you think is the difference between combinations and permutations

[willow bear]

"combinatorics" is the name for the entire field of math

nCr is called "combinations"

[valid violet]

The mnemonic i was taught is

Combination locks use permutations

@odd helm

[harsh cipher]

hi

what log rule is this

\log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)

@obsidian monolith \log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)

[valid violet]

dollar signs

[harsh cipher]

$\log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)

$ \log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)

[valid violet]

both sides

[harsh cipher]

$\log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)$

[obsidian monolith]

אewb64:

[valid violet]

write $\log_{a^b}(x) = \frac {\ln(x)}{\ln(a^b)}$

[obsidian monolith]

gfauxpas:

[harsh cipher]

natural logarithm?

[valid violet]

or any other logarithm if you'd like

$\frac{\log_2(x)}{\log_2(a^b)}$

also works

[obsidian monolith]

gfauxpas:

[valid violet]

doesn't matter

[harsh cipher]

im trying to solve this problem

and I don't understand the section where it has the light bulb icon

https://www.symbolab.com/solver/factor-calculator/2^{\log_{8}\left(x\right)^{27}}

[valid violet]

I'm telling you where the rule comes from

pic a logarithm and do a change of base of logarithm

$\log_x(y) = \frac{\log_{10}(y)}{\log_{10}(x)} = \frac{\log_{e}(y)}{\log_{e}(x)} = \frac{\log_{2}(y)}{\log_{2}(x)}$

[obsidian monolith]

gfauxpas:

[harsh cipher]

okay...

[valid violet]

$\log_{a^b}(x) = ?$

[obsidian monolith]

gfauxpas:

[harsh cipher]

$\log_{b/a}=a$

[obsidian monolith]

אewb64:

[valid violet]

[harsh cipher]

lol

hahahaha

[valid violet]

use the rule I told you to use

[harsh cipher]

change of base

[valid violet]

yes

pick a base

and apply it to log_{a^b}(x)

look,

[harsh cipher]

listening

[valid violet]

$\log_y(x) = \frac{\log_{2}(x)}{\log_{2}(y)}$

[obsidian monolith]

gfauxpas:

[valid violet]

agree?

[harsh cipher]

totally yes

[valid violet]

now try this with y = a^b

[harsh cipher]

log y / log a?

[valid violet]

where'd the b go

just take the equation I typed and everywhere you see "y" put in "a^b"

[harsh cipher]

I need to redo my lesson from changing to exponential equation to log form

[valid violet]

we're not using the exponential equation at all

[harsh cipher]

y= a^b is in exponential form?

[valid violet]

$\log_y(x) = \frac{\log_{2}(x)}{\log_{2}(y)}$

err backwards

[harsh cipher]

yeah

[obsidian monolith]

gfauxpas:

[valid violet]

every time you see the symbol "y" in this equation

take your mental eraser and scrub out the "y"

and instead write in "a^b"

$\log_{\phantom{y}}(x) = \frac{\log_{2}(x)}{\log_{2}(\phantom{y})}$

[obsidian monolith]

gfauxpas:

[valid violet]

I erased the "y" for you

now write in a^b every place there used to be a y

[harsh cipher]

k

$\log{{b}|(x)= \frac{\log{2}(x)}{\log{2}(b)$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

shoot

[valid violet]

$\log_{b}|(x)= \frac{\log{2}(x)}{\log{2}(b)}$

[obsidian monolith]

gfauxpas:

[valid violet]

uhh

whateve ryou were trying to write

look

$\log_{a^b}(x) = \frac{\log_{2}(x)}{\log_{2}(a^b)}$

[obsidian monolith]

gfauxpas:

[valid violet]

do you see why this is true

[harsh cipher]

it's the change of base rule

[valid violet]

yes

but you were having a hard time writing it

I'm not sure why

[harsh cipher]

its like trying to solve 14 = 2^x

[valid violet]

no

I'm not trying to solve anything

I am just trying to rewrite the logs as logs with a different base

$\log_{r}(s) = \frac{\log_{2}(s)}{\log_2(r)}$

agree or disagree?

err its backwards again

[harsh cipher]

agree

s/r

[obsidian monolith]

gfauxpas:

[valid violet]

okay

[harsh cipher]

how does that relate to my question

[valid violet]

is this equality still true if I rename "s" to "x"?

[harsh cipher]

yea

[valid violet]

is it true if I rename "r" to "a^b"?

[harsh cipher]

yea

[valid violet]

so do you agree that the expression you're trying to analyze,

$\log_{a^b}(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

is equal in every way to the expression

$\frac{\log_2(x)}{\log_2(a^b)}$

[obsidian monolith]

gfauxpas:

[valid violet]

?

[harsh cipher]

kind of...

[valid violet]

until now you agreed that you can rename the variables anything I'd like

s to x, and r to a^b

[harsh cipher]

yea

[valid violet]

so why "kind of"?

[harsh cipher]

well

$\log _{a^b}\left(x\right)=\frac{1}{b}\log _a\left(x\right)$

[obsidian monolith]

אewb64:

[valid violet]

what about it

[harsh cipher]

its multiplying 1/b?

[valid violet]

I didn't get there yet

because you are having trouble first with the change of base formula

and I need to get that worked out first

[harsh cipher]

uh

would you call this reciprocal law for logarithm?

cuz that's what I learned in my online lessons

[valid violet]

$\log_{a^b}(x)=\frac{\log_2(x)}{\log_2(a^b)}$

[obsidian monolith]

gfauxpas:

[harsh cipher]

-_-

[valid violet]

I cannot help you until you are 100% convinced of this equality

[harsh cipher]

I got it

[valid violet]

not just kind of

are you convinced?

[harsh cipher]

yea

[valid violet]

$\

$\log_2(a^b) = b \log_2(a)$

[obsidian monolith]

gfauxpas:

[harsh cipher]

if you change the letters or symbols I can explain to you now

[valid violet]

and that's my hint

you can finish the proof

[harsh cipher]

=1?

[valid violet]

$\log_{a^b}(x)=\frac{\log_2(x)}{\log_2(a^b)}\text{ and }\log_2(a^b) = b \log_2(a)$

[obsidian monolith]

gfauxpas:

[harsh cipher]

log x / log a^b

you take the log a^b / log base 2

[valid violet]

$y=\frac xz \text{ and } z = t , \text { then } y=\frac xt$

[harsh cipher]

I get that

[obsidian monolith]

gfauxpas:

[harsh cipher]

Ok

[valid violet]

I cannot give you any more hints without telling you the answer

$\log_{a^b}(x)=\frac{\log_2(x)}{\log_2(a^b)}\text{ and }\log_2(a^b) = b \log_2(a)$

[obsidian monolith]

gfauxpas:

[harsh cipher]

Ok first part makes sense

ok power rule in reverse for second equation

[obsidian monolith]

gfauxpas:

gfauxpas:

[valid violet]

[harsh cipher]

what do you want me to write?

top part is supposed to be

log base (x) should be log a^b

(x)

not log base (x)

[valid violet]

I'm sorry i don't know how to explain it any more than what i did.

Try taking a 10 minute break then reading it again

[harsh cipher]

okay I'll be back

😦

[agile steppe]

For number 3, shouldn’t it be 1-2A^2 ?

It says 1-A^2 (I don’t know if I did something wrong, but I still think it is 1-2A^2

I mean, I remember the identity was 1-2sin^2theta

Please tell me if I am wrong

[valid violet]

What is A

@agile steppe

[willow bear]

it says in the photo

sin(x) = A

[valid violet]

Yes im asking him

[harsh cipher]

Ann

[willow bear]

anyway... he's right

[harsh cipher]

were you here the whole time?

[willow bear]

no

[harsh cipher]

hahah okay

[valid violet]

Ann maybe you can help me help newb

[willow bear]

@agile steppe yes it should be 1 - 2A^2

@valid violet i'd rather not. i'm having a bit of a pain spike and i'm also kinda tired

[harsh cipher]

hahahaha

[valid violet]

Okay np

Newb

[harsh cipher]

yea

[valid violet]

Let y = log_a^b(x)

r=log_2(x)

s=log_2(a^b)

t=b log_2(a)

Do you agree that y = r/s

[harsh cipher]

No

[valid violet]

@agile steppe sorry coldstein i took the question at face value, their identity is false