help-23

[safe radish]

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[half cliff]

np

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[limpid cape]

hey

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@rigid kernel Has your question been resolved?

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[dusky torrent]

How to do Fourier analysis?

[dusky torrent]

There is something like this i heard

Actually what is it ?

[wind onyx]

This belongs in ode and pde channel. There are good videos online that cover signal processing which deals with former transforms. I suggest u check it out as it’s an important application

[safe radish]

@dusky torrent Has your question been resolved?

[light shoal]

this question is way too broad, could be the subject of many books

[dusky torrent]

I wanna learn by a human being

Practically

[plucky elk]

Go hire a tutor

[dusky torrent]

I mean just practical explanation

[plucky elk]

Help channels are for specific problems

[dusky torrent]

To learn something

[plucky elk]

Then Google it

[dusky torrent]

Tried couldn't understand

Why we even use Fourier analysis

[plucky elk]

Your query is too vague like bungo said

[dusky torrent]

Just tell me that

[plucky elk]

Google applications of Fourier series

[safe radish]

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[mossy ivy]

can someone explain to me how i’m supposed to graph this?

[limpid cape]

hey

i’m back

[mossy ivy]

yo

i was gonna dm you but i didn’t wanna feel like a bother to you 😭

[limpid cape]

ur all g

anyways

let’s first solve for x

lmk when you got it or if you need help finding it

[mossy ivy]

x can be any number tho right

[limpid cape]

yes

wait

wdym?

solving for x

[limpid cape]

oh damn what’s this

[mossy ivy]

oh wait is this the wrong one

oh na that’s the right one

those are the numbers i chose

[limpid cape]

why are you choosing numbers

[mossy ivy]

for x

[limpid cape]

we need the vertex, and x values to sketch the graph

x^2 = 3

[mossy ivy]

fuck so i started it hella wrong then huh

[limpid cape]

the answer is sqrt(3) and -sqrt(3)

those are your X’s

[mossy ivy]

wait how you get sqrt(3) for x

[limpid cape]

ohhhh wait

[limpid cape]

you’re manually plotting?

[mossy ivy]

yessir

[limpid cape]

shit my fault

[mossy ivy]

you confused me lmao but it’s all good

[limpid cape]

just plot those numbers

left side is your x, right side is your y

so (-3, 6)

and so on

[mossy ivy]

am i doing this right lol

[royal pivot]

no

[mossy ivy]

fuck

[limpid cape]

yea that’s not correct

[mossy ivy]

(-3,6) (-2,1)

[royal pivot]

you have to plot it

input values of x and output values of x²-3

[limpid cape]

your vertex is (0, -3)

[mossy ivy]

i did -3^2= 3x3=9-3=6

that’s how i got (-3,6)

[limpid cape]

yea but that’s not your vertex

[royal pivot]

that is a point tho

[limpid cape]

this shows 3, -6 as the vertex

which isn’t correct

[royal pivot]

plot (0.-3) on too

[limpid cape]

plot that first

[royal pivot]

you need 3 points for that graphing software

[limpid cape]

oh perfect

remember the square roots i was talking about?

[limpid cape]

^

you need your vertex and 2 points

[limpid cape]

first plug sqrt(3) into the original equation

[limpid cape]

one moment, i just spilled my root beer everywhere

alright let’s continue

so @mossy ivy

[limpid cape]

this answer will be your y value

[lean otter]

Do they even know how you got sqrt(3) ?

[limpid cape]

i mean x^2 = 3

i jumped straight to the answer

[lean otter]

why?

[limpid cape]

i don’t see how this diagram helps

to cancel out the x^2, you’d square both sides @mossy ivy

so sqrt(x^2) and sqrt(3)

[lean otter]

x^2-3 are two squares

if you know the difference of squares theorem is

you would know the zeros would be plus plus or minus sqrt(3)

it also helps you factor the equation

[limpid cape]

my apologies then

i assumed he knew the algebra

or she

[mossy ivy]

bruh im lost lol i feel stupid like i should know this but ima take a break lol cuz my brain is fried man

i can’t think straight

[limpid cape]

i think we need to go back to basic algebra first bro

lmk if you’d like some help, def take a break tho

i’ve been helping you for the past hour and half

[mossy ivy]

alright

thank y’all so much

[lean otter]

next time remeber that you don't actually have to plug values into the function.
The two images i shared are all you need to solve the problem

[limpid cape]

lmk when you’re ready to learn again

[mossy ivy]

@lean otter okay gotchu

@limpid cape and i will thank you for being patient

[limpid cape]

no problem

[safe radish]

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[neat onyx]

How is 3d solved

[neat onyx]

Sorry 3c

I don't know ehat to do cause I' don't have the value for mu or standard deviation

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@neat onyx Has your question been resolved?

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@neat onyx Has your question been resolved?

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[pale birch]

lemme take the pic again 1 sec

^what the correct answer should be

I have been searching for a while to see what I did wrong but can't find it

[cosmic grove]

$(7+11i)² -4(2+i)(1+38i) = 49 + 154i -121 -4(2+39i-38)$

[flat frigate]

Herels

[pale birch]

Could you explain to me as to why am not supposed to take the '-' before b?

[cosmic grove]

?

[pale birch]

why 7+11i and not -7-11i

^2

[cosmic grove]

thats the same thing tho

(-7-11i)^2 = (-1)^2 (7+11i)^2 = (7+11i)^2

[pale birch]

oh so that wasnt the problem

I think I can continue now, thank you for your time & effort

.close

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[exotic cipher]

how to solve this question? both a and b

[exotic cipher]

<@&286206848099549185>

[royal kiln]

You need to make a chart

Keep in mind

That the chart contains

[uncut oasis]

create a tree diagram

[royal kiln]

The intersections of probabilities

But you are given conditional probabilities

So there is a tiny bit of Algebra as well as arithmetic to this

P(pass=T|studied=T) =0.9
P(pass=T|studied=F) =0.4
P(studied=T) =0.8

You need to know that

P(A|B) = P(A and B)/P(B)

Fill this in

[exotic cipher]

what does =T and =F mean

[royal kiln]

Again, the 4 upper-left elements are intersections (AND's), not conditionals. Bottom-right entry is 1.0 (forgot to write it)

[royal kiln]

Shorthand for is true, is false

[exotic cipher]

the first column would total greater than 1 tho

.4 + .9

[royal kiln]

I'm kind of wondering if I misinterpreted something b.c. while my graph's arithmetic adds up, the answers are the same as what's given to you, and that can't be what they intended

I don't have the same column entrees as you

I guess what's given to you are intersecting events and not conditionals?

I started with that assumption and had the same problem you are having

I think this working is mistaken ^

[royal kiln]

Do you know where I went wrong here?

At the start, or in the middle?

[safe radish]

@exotic cipher Has your question been resolved?

[uncut oasis]

no, in all honesty it looks right to me

[exotic cipher]

yeah i can leave it as that

thanks

.close

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[past sage]

stuck on part a

it says take a = T(1,0,0)

so do i just take b to be T(0,1,0)

and c similarly

?

if so what comes next

[lean otter]

yes, those are good choices

[past sage]

i'm stuck on what to do after tho

[peak estuary]

well that's it

do you know what a matrix representation of a linear map is?

[past sage]

not yet

i think we're going over matrixes nec

next

but it's weird this other dude who was helping me just proved T was linear

but i thought it was given that it was linear

[peak estuary]

it is

[past sage]

he didn't even take a to be T(1,0,0)

[peak estuary]

when you know what matrix representations of linear maps are, you know that any linear map R^n -> R^m can be presented by an mxn matrix

which in this case would be 1x3 (a,b,c)

[past sage]

okay hang on

so for this question my answer will be then

Take a to be .... b to be .... c to be ..... Thus there exists real a,b,c such that T(x,y,z) = ax+by=cz

this just seems too short to be an answer

[peak estuary]

well you have to show that it indeed works

using the linearity of T

[past sage]

oh so after writing that

then I have to prove T is linear

using two vectors

x,y,z and x',y',z'

[peak estuary]

no

T is linear

that is given

you have to show that the equality T(x,y,z)=ax+by+cz is true

[past sage]

so then after defining a,b and c i just write now take T(x,y,z) = T(1x,1,y,1z)

[peak estuary]

no

[past sage]

=ax+by=cz

mhmmm

darn i'm not sure how to get there

[peak estuary]

you used the unit vectors to define a,b,c

so you should somehow get from T(x,y,z) to T(1,0,0) etc

using linearity

[past sage]

T(x,y,z) = xT(1,0,0) + yT(0,1,0)....

like this?

[peak estuary]

yes

[past sage]

ohhh

then i say that equals to =ax+by+cz

and we are done

[peak estuary]

yes

[past sage]

thank you for helping and not giving me the answer

i needed a little push to remember linearity rules

[peak estuary]

thats the point of this server

[past sage]

hehehe

hey for part b he mentioned that the equations are all the vectors of R^n right

or are those the vectors of W

[peak estuary]

what

W is not all of R^n

eg (1,1,1...,1) is clearly not in W

[past sage]

i'm a bit confused by what he means when he says W is the subspace of

R^n

of equation x1+....xn=0

[peak estuary]

W is the set of all (x1,...,xn) in R^n which satisfy x1+...+xn=0

[past sage]

wait

if i'm not mistaken

this is kind of like a linearmapping from V to R^n

where W is the Range(T)

since rangeT is a subspace of the codomain right

Wait forget that

since W is an equation of n vectors =0

then we can take -xn=x1+....x_n-1

so W has a dimension n-1?

[safe radish]

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[plush depot]

if it's dy/dx why is the result dx/dy?

[hardy lion]

what do you mean the result is dx/dy?

[worthy hemlock]

The result isn't dx/dy

It's still dy/dx

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[west belfry]

How do i factor x^2-1? I'm confused on i stuff

[lean thorn]

difference of squares. You use the following: $a^2 - b^2 = (a + b)(a - b)$

[flat frigate]

MellowDramaLlama

[shrewd geyser]

A great method to factor polynomials is the cross method. I'm sure there are videos online that can go through some examples

[west belfry]

so is it x-sqrt1 and x+sqrt1?

[lean thorn]

yep but what is the sqrt(1)?

[west belfry]

just 1

oh my goodness

[lean thorn]

bingo

(x + 1)(x - 1)

[west belfry]

ok thanks

[lean thorn]

and if you FOIL that out you'll see it matches

[west belfry]

yes,

how about x^2+4?

i was trying to find a question about using i

[obtuse plover]

This was excellent

[shrewd geyser]

x^2 + 4 cannot be factorable using real numbers

[west belfry]

yes I need to use imaginary but im confused

[shrewd geyser]

You can do it using complex numbers

[west belfry]

yes!

[shrewd geyser]

You can use the difference of squares formula again

If you use this along with complex numbers you will be able to factor it out

[west belfry]

so it's x+2i and x-2i?

[shrewd geyser]

bingo bongo bango

you're a genius

[west belfry]

thanks!

I appreciate it

.close

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[mossy wadi]

im stuck on this last problem

[crimson field]

x = ln(t), y = π/2 t. You can then find z = f(x(t), y(t)) then you have r(t) = <x(t), y(t), z(t) and you can deploy the usual tactics. Clearly the point P corresponds to t = 1.

[safe radish]

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[crimson field]

You're welcome! @mossy wadi

[mossy wadi]

:D

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[frosty granite]

using part a i got 2sin4xcos3x=0

[frosty granite]

if thats right how would i solve that lol

[rotund wolf]

Divide the two over

[frosty granite]

oh

wait u just split it?

sin4x = 0 etc?

[rotund wolf]

And then consider what makes sin4x 0

[frosty granite]

thats kinda silly

but htanks

[rotund wolf]

Yeah

[frosty granite]

thanks

[rotund wolf]

Nw

[frosty granite]

.close

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[lean otter]

Am I correct in saying continuity implies differentiability and differentiability does not imply continuity

[royal pivot]

i think it’s the other way around

nvm

[tiny wraith]

It is

[lean otter]

How would I show a graph is differentiable in a certain interval?

[tiny wraith]

Probably take the derivative and show it exists everywhere on the interval

If the derivative exists then the function must also be continuous

[lean otter]

Okay thank you

so diff -> continuity but continuity does not -> diff

[tiny wraith]

Yes

[lean otter]

okay perfect thanks

.close

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[royal pivot]

but you can differentiate 1/x

[lean otter]

.reopen

[safe radish]

✅

[tiny wraith]

Not at 0

It isn't even continuous there

[lean otter]

and also for |x| it isnt differentiable at 0

even though its continuous

[royal pivot]

we already established that

cont does not imply diff

[tiny wraith]

Yup

[lean otter]

yeah I know

.close

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[lean otter]

.reopen

[safe radish]

✅

[lean otter]

is the notation f'(f^-1(a)) just the derivative of the inverse of f(a)?

[tiny wraith]

It is the derivative evaluated at (the inverse of f at x)

[viral loom]

evaluated at a

[tiny wraith]

Oh, yes

[lean otter]

Oh right! The correct notation for the derivative of the inverse of f(a) is actually (f^-1)'(a) right?

[tiny wraith]

Yes, that would make it clear

[lean otter]

and that would equal 1 / f'(f^-1(a))?

[tiny wraith]

Yes

[safe radish]

@lean otter Has your question been resolved?

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[gritty umbra]

$y=2x³cosx-3x²secx+5sinx$

[flat frigate]

Sterling

[sonic geyser]

?

[gritty umbra]

derivatives of Trigonometric functions

sorry for the lack of context I tried solving it

[viral loom]

what is that?

[safe radish]

@gritty umbra Has your question been resolved?

[gritty umbra]

yeah I'm not sure if I did that right so I asked here

I though the derivative of 5 is always 0

[viral loom]

it's multiplying sinx

[gritty umbra]

so 0 multiplied by the derivative of sinx is 0

[viral loom]

what's the derivative of 5x?

[gritty umbra]

5

but the given is 5 only, which is constant

or is it x is my sinx

[viral loom]

use the product rule to understand why derivative of 5x = 5

and then use the product rule for 5sinx

[gritty umbra]

I don't understand where you got the 5x from

the given is 5sinx

[viral loom]

it was an example

[gritty umbra]

oh ok

[viral loom]

if dev 5x = 5

not 0

even derivative of 5 is 0

because you use the product rule

[gritty umbra]

I don't need the product rule for 5x tho

[safe radish]

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[gritty umbra]

it's just 5

.reopen

[safe radish]

✅

[viral loom]

i just wanted you to know why derivative of 5x is 5, so that way

[gritty umbra]

so what do I do now?

[viral loom]

you can understand why

[gritty umbra]

okay

[viral loom]

derivative of 5sinx is not 0

it's 5cosx

[gritty umbra]

why?

product rule?

[viral loom]

derivative of 5 = 0

5sinx = 0

derivative of sinx = cosx

5cosx

[gritty umbra]

oh

it's 5cosx + 0

[viral loom]

0 + 5cosc

yes

the same for 5x

it's 5 + 0

it's the same

but we do right away because it's trivial

[gritty umbra]

,Wolfram d/dx(2x³cosx-3x²secx+5sinx)

[flat frigate]

Results provided by WolframAlpha

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

Requested by Sterling#7883

[gritty umbra]

um

I got

$-2x³sinx+(6x²+5cosx)-3x²secxtanx-6xsecx$

[flat frigate]

Sterling

[gritty umbra]

@viral loom is this correct?

please

[viral loom]

there is a mistake

this one is not correct

(a+b)c = ac+bc, not a +bc

[gritty umbra]

right, is it (6x²+5)cosx

[safe radish]

@gritty umbra Has your question been resolved?

[gritty umbra]

<@&286206848099549185>

[safe radish]

@gritty umbra Has your question been resolved?

[gritty umbra]

.close

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[sudden sequoia]

How can I find the radius of convergence for this?

[tiny wraith]

You could compare it with a geometric series

[sudden sequoia]

Could you elaborate a bit more?

[tiny wraith]

Okay so do you know when the sum of z^k converges, without the sin(k)

[sudden sequoia]

When z is 0<z<=1 ?

[tiny wraith]

z can be between 0 and -1 as well

So -1 < z < 1

[sudden sequoia]

Hmm alright

[tiny wraith]

So, in this case, if z <= -1 or z >= 1, then the series converges

Because the terms don't go to 0

If -1 < z < 1, then you can compare this series with the geometric series, because -1 <= sin(k) <= 1

[safe radish]

@sudden sequoia Has your question been resolved?

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[slow cove]

I used MathWay and other applications, but they keep saying it's 5 sqrt(3) /6 - 2. I don't get why

[thin bridge]

they split the fraction
or didn't even bother combining the -2

[prime holly]

yeah

[thin bridge]

what was the question asking for

[slow cove]

Well this is from trigonometric values of special angles and I need to simplify them

[thin bridge]

either result is fine

[slow cove]

Okay thank you

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[hot thistle]

I don't see how u' satisfies the equation when substituted for g

[hot thistle]

ibp would suggest the sum of the integrals to be u * phi, which I'm not sure would necessarily be 0

nevermind that's a boundary term that goes to 0

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[lean otter]

Could the answer for this be 10,000?

The integer that would allow data a to have a larger mean would be m is greater than 42

So it has to be m is greater than or equal to 43

And the largest value in the set is 46 so

Would the answer be anything 46 or greater?

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[lean otter]

.close

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[safe radish]

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[wanton canopy]

.close

.close

?

.reopen

I would like to get some help, our professor gave us an integral to try and solve which was:
Integrate\int_{ }^{ }\cos\left(\ln\left(\sin\left(x\right)\right)\right)dx , a really hard one in fact. He told us that we can ask for help online and look up how we could solve it, I tried my best to solve it and i ended up with \int_{ }^{ }\cos\left(\theta\right)^{i}d\theta, I was wondering if it's possible to solve?

im not sure how to use the latex bot ill just take sc of demos

the problem to solve

what i ended up with

Excuse my bad handwriting 😅

also i only took calc 2 so far, i dont know much about complex integration

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[wanton canopy]

Okay ill just resend the question:
I would like to get some help, our professor gave us an integral to try and solve which was:
Integrate\int_{ }^{ }\cos\left(\ln\left(\sin\left(x\right)\right)\right)dx , a really hard one in fact. He told us that we can ask for help online and look up how we could solve it, I tried my best to solve it and i ended up with \int_{ }^{ }\cos\left(\theta\right)^{i}d\theta, I was wondering if it's possible to solve?

[wanton canopy]

The problem

[wanton canopy]

My work

[safe radish]

@wanton canopy Has your question been resolved?

[empty gyro]

Use dollar signs: $\sqrt{x}$

[flat frigate]

SWR

[wanton canopy]

ohh mb

$Integrate\int{ }^{ }\cos\left(\ln\left(\sin\left(x\right)\right)\right)dx , a really hard one in fact. He told us that we can ask for help online and look up how we could solve it, I tried my best to solve it and i ended up with \int{ }^{ }\cos\left(\theta\right)^{i}d\theta, I was wondering if it's possible to solve?$

[flat frigate]

Lunar

[wanton canopy]

oof lol wait

[empty gyro]

Surround only your math terms in dollar signs.

[wanton canopy]

Integrate$\int{ }^{ }\cos\left(\ln\left(\sin\left(x\right)\right)\right)dx$

[flat frigate]

Lunar

[empty gyro]

I do not believe it has a simple closed form

,w integral cos(ln(sin(x))

[flat frigate]

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[empty gyro]

Yeah, you gotta use hypergeoemtric function.

[wanton canopy]

hm does that mean its non elementary?

[empty gyro]

yes

[wanton canopy]

ah okay thank you! ig there is no way to solve it unless it was an indefinite integral?

or like

with bounds

[empty gyro]

with bounds maybe? But hard to say

[winter pivot]

DEFINITE integrals have bounds

[wanton canopy]

Thank you for your help then :)

.close

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[lean otter]

apparently this is an Olympiad question: [
x^{x^{4}} = 64
]

[lean otter]

how would you go about solving for x?

[hazy elbow]

Raise both sides to the power 4

[pulsar pecan]

I would sub in numbers and see if they work

[main ice]

no beard is right

don’t do that ^

[hazy elbow]

That's hit and trial

[main ice]

it won’t work for this problem

[pulsar pecan]

it is easy to do as a first step

[hazy elbow]

The solution is pretty complex and so pretty hard to guess

[hot thistle]

this is nice

[swift lily]

Newton rhapson

x^x^4 - 64 = 0

[peak estuary]

very nice

[swift lily]

Find root ig?

[hazy elbow]

Pretty complex

[swift lily]

Not rlly if you're familiar with calc 1

[hazy elbow]

Raising both sides to power for and a substitution makes this probelm a piece of cake

[vast cloud]

^ yep

[peak estuary]

you are supposed to do this by hand

and its entirely possible with what beard wrote

[hot thistle]

how would you do this using calculus. i can't see it

[hazy elbow]

Don't even need that stuff

[swift lily]

@hot thistle

[peak estuary]

newton would not find the closed form

[pulsar pecan]

$$ (x^{x^4})^4 = 64^4 $$

[flat frigate]

l33t_syncopations

[hot thistle]

that's not an exact solution

[swift lily]

True

[pulsar pecan]

$$ x^{4x^4} = 2^{24} = 8^8 $$

[hot thistle]

this is math right? we only care about existence of solutions. use IVT

[pulsar pecan]

is this right?

[main ice]

no

[hazy elbow]

Then try to writw 64^4 in the form a^a

[main ice]

right side becomes 8^8

[hazy elbow]

Yes

[flat frigate]

l33t_syncopations

[hazy elbow]

Yes

Now assume x^4=a

[pulsar pecan]

oh, $$ {x^4}^{x^4} = 8^8 $$

[flat frigate]

l33t_syncopations

[pulsar pecan]

but is there another solution other than 2^{3/4}

[hazy elbow]

x^4=8

4 solutions (including complex)

[hot thistle]

make sure they aren't extraneous from raising to the fourth though

[pulsar pecan]

and other than those?

[hazy elbow]

,w solve x^4=8

[flat frigate]

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[swift lily]

[pulsar pecan]

oh

and how are you supposed to come up with the initial trick of raising to the fourth

[swift lily]

Use your brains

[hazy elbow]

Intuition

[hot thistle]

if it's an olympiad question, then it's more than likely a well known trick

[hazy elbow]

Yess

Especially when base and power both contain a variable

[hot thistle]

i wonder if you can also solve this using a taylor polynomial

[swift lily]

Lmaooo

Pain

[lean otter]

oh um

did not think this channel would get this active gee

but raising to the power of 4 thats smart

[pulsar pecan]

would that not be inexact

[swift lily]

Fr

Usually people would think it'll make it worse

[hot thistle]

how so

[pulsar pecan]

but it’s probably like the first time you see a fork in chess

[swift lily]

it's because your name has 2 hearts next to it

[pulsar pecan]

how do you use a taylor polynomial to solve this

[lean otter]

just curious, how did u see it? @hazy elbow

[hot thistle]

i just said i wonder if it's possible

[swift lily]

Intuition™️

[lean otter]

are you just familiar with Olympiad questions?

[hazy elbow]

I have solved these type of problems earlier. Also it seems kinda obvious when the base and power contain same variable

Quick one
Solve this
15x^3x=5

[swift lily]

Nani

[lean otter]

[
(15x)^{3x}=5
]?

[hazy elbow]

Yes

[swift lily]

dies

[lean otter]

is it like

[swift lily]

Wait is (15x)^3x or 15 (x)^3x

[hot thistle]

assume the latter unless you have parentheses

[hazy elbow]

This

(15x)^(3x)=5

[swift lily]

Oh so it's the ENTIRE THING that's raised to the power

Kk

[pulsar pecan]

would that be dividing by 5 on both sides

[hot thistle]

oh now i'm confused

[vast cloud]

||raise to power 5||

[swift lily]

Dw same

[hazy elbow]

Close but not enough

[lean otter]

i assume the same logic then

just raise to the power 5

ta da

[swift lily]

Lmaoooo

Hacks fr

[hot thistle]

if the 15 wasn't raised, this should be correct

[swift lily]

Bow wharf is 15 wasn't?

Now what if 15 wasn't?

[lean otter]

wait is it always true that [
a^a = b^b \implies a = b
]

[hazy elbow]

Yes

[swift lily]

Hm

[pulsar pecan]

Why

[swift lily]

Yeah should be

Ok time to prove it

[hazy elbow]

This may seem incorrect but try putting values

[swift lily]

Ok ima try proving

Man why am I just giving myself more work to do

[lean otter]

lmaoo

[hot thistle]

x^x isn't 1-1

[lean otter]

what even is x^x

,w graph x^x from -10 to 10

[flat frigate]

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[swift lily]

Oh that

[hot thistle]

don't go to 10

[swift lily]

Irs a weird graph

[hot thistle]

it blows up too quick

[swift lily]

Ye

[hot thistle]

,w graph x^x from -1 to 2

[lean otter]

oh yeah damn

[swift lily]

I mean expected

LOL

Just desmos it

[flat frigate]

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[lean otter]

interesting it plummets a bit

[hazy elbow]

Oh so it is not true

[swift lily]

Wait it's not defined for negative inputs?

[hot thistle]

this is the infamous 0^0 = 1

[hot thistle]

well it's not properly defined for x = 0, so they probably just cut the domain there

[swift lily]

LOL

[pulsar pecan]

0^0 = 0/0 = sin 0 / tan 0 = cos 0 = 1

[lean otter]

but like -2^-2 is a thing no?

[hot thistle]

also, negative^x is tough to parse for all reals x

[hazy elbow]

Probably cuz (-1/2)^(-1/2) is imagniary

[swift lily]

Desmos says no

[vast cloud]

No

[hot thistle]

it'd just be -1/4

[vast cloud]

½^½ = ¼^¼

for example

[hot thistle]

or 1/4

since you square haha

[swift lily]

Damn

[hot thistle]

but point is (negative)^irrational is undefined

[hazy elbow]

Okay for x>1 it is valid

[vast cloud]

Yes

[hot thistle]

and also undefined for a lot of rationals

[lean otter]

hmm i see

[lean otter]

oh yeah it just doesnt work for any number lol

1^1 = 2^2

[hot thistle]

you need b^b >= 1 for unique solutions (avoiding a = 0)

[swift lily]

It isn't?

[lean otter]

wait so how can u verify the question's solution

[hot thistle]

how would you define it

[vast cloud]

like beard said tho if x,y in [1,inf) then x^x=y^y implies x=y

[hot thistle]

you're going to get complex solutions

[swift lily]

||deez nuts?|| wait WHAT

My brain cannot comprehend

[lean otter]

oh my god harper u r a genius i was staring at this for 5 minutes trying to figure out why this is even true

[swift lily]

Also me:
plugs into calculator

"Checks out?"

"Yeah."

"OK works"

[hot thistle]

in general, exponents such as 2^pi are defined in terms of limits, as in, for a sequence (a_n) that tends to pi, we define 2^pi as the limit of 2^(a_n) as n goes to infinity
this limit is not unique for negative bases

that is, the limit doesn't exist

a simple example is
$$\lim_{n\to\infty}(-2)^{\frac{1}{2^n}} \ne \lim_{n\to\infty}(-2)^{\frac{1}{3^n}}$$

[flat frigate]

maximo

[hot thistle]

oh that one might actually work

they might even work in general

[swift lily]

LOL

[hot thistle]

ok no it shouldn't

one gives you 1, the other -1

this also goes without saying, but we would be allowing complex numbers which, depending on context, also makes this no good

[swift lily]

Bruh smth's wrong with my net

[hot thistle]

i think its discord

[swift lily]

Aaaanw yeah it doesn't..
Ic

Lmao convergeence

Ok bruh proving that a^a = b^b assuming a,b>1 is ummmm hard

Why am I doing this to myself

[hot thistle]

you "just" need to show x^x is 1 to 1

but if you know calc it shouldn't be difficult

[hazy elbow]

will it like require the Lambert W function?

[hot thistle]

nah

pf: ||x^x is strictly increasing for x>1, so it is one to one||

[swift lily]

Hmm

AHHHHH

WHY DID I UNSPOILER THAT

[hot thistle]

i put the spoiler on purpose

[swift lily]

[swift lily]

[keen comet]

The actual inverse will need this I think

[swift lily]

Lmao I like how lix basically vanished from here

[swift lily]

Yee

[keen comet]

I'm of course assuming you have done the restriction

[swift lily]

"Solve lambert W"

Sure, ,w

[hot thistle]

my favorite inverses are those for non-injective functions

[swift lily]

[safe radish]

@lean otter Has your question been resolved?

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[untold topaz]

Is this how u do A?

[main ice]

you can use desmos to check

[untold topaz]

Bro I don't even know if my working is correct

What more to say input into desmos 🥲

My first time seeing (a) btw

[main ice]

looks right to me but double check with desmos

[untold topaz]

x=/ 2
?

[safe radish]

@untold topaz Has your question been resolved?

[safe radish]

@untold topaz Has your question been resolved?

[safe radish]

@untold topaz Has your question been resolved?

[hushed minnow]

What does that mean?

@untold topaz

[untold topaz]

It's just to help me see the mapping rule

[hushed minnow]

looks good

[untold topaz]

.close

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[stiff aurora]

what is inverse of 1by1 matrix

[native tartan]

inverse of a number?

so just 1/n

[stiff aurora]

cant be that

[native tartan]

?

why not

[stiff aurora]

the value is inside a matrix

[main ice]

it is 1/n

[native tartan]

$\begin{bmatrix}n\end{bmatrix}^{-1}$

[flat frigate]

Poelymole

[stiff aurora]

can u illsutrate by taking 'a' in matrix

and give its inverse

and tell its cofactor also

[native tartan]

$\begin{bmatrix}a\end{bmatrix}^{-1} = \frac{1}{a}$ ?

[flat frigate]

Poelymole

[native tartan]

im confused about what you mean

[stiff aurora]

when we find adjoint we find cofactors first by deleting nth row and column associated with that given element

do that same thing here and tell me cofactor

[native tartan]

but theres only 1 row and column?

so cofactor is 0?

im guessing it just doesnt have one tbh

[stiff aurora]

see that is the ques of this day

thats y im confused kind of

[native tartan]

yeah man i think ur on ur own here

pushing the boundaries of maths fr

[stiff aurora]

might say that but can be some hinderance in knowledge of me

gonna close and ask aagain

ty for concern anyway

close..close

.close

[safe radish]

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[lean otter]

Does the last statement implication go both ways or only forwards

[safe radish]

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@lean otter Has your question been resolved?

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[bleak scroll]

{u1,u2} is a base

and f:v -->V is a linear map

these are f operations

calculate

@granite idol

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[true estuary]

for part a do i need to paramterise x and y in terms of sine and cosine?

or could i do it directly with x and y

cuz im trying to do it directly in terms of x and dx but then would my limits of x be 0 and 0?

surely that doesnt make sense tho since id get 0

[final halo]

you always need to parametrise when you're doing a line integral

thats kind of the definition

[true estuary]

what if i paramterised y in terms of x

cuz i tried that

ignore my limits

[final halo]

your x limits would be from -1 to 1

[true estuary]

why tho? the limits its from (0,-1) to (0,1) and 1 and -1 are y coordinates?

[final halo]

you have parametrised by (x,sqrt(1-x^2))

just imagine those points in the plane

the left most point of the semi circle is when x=-1, the rightmost is when x=1

this^

[true estuary]

if i paremterised in terms of cos and sine

would my limits be 0 and pi?

[final halo]

yea

[true estuary]

i picked between 0 and pi and got an answer of 0

idk if i made a mistake

cuz the answer should be 2/3

[final halo]

oh misreading the Q

its the right half of a circle

[final halo]

you were right here i was mistaken

which means if you wanted to do your original way, you would parametrise by y instead

[true estuary]

yh made sense

but in terms of sine and cosine

why are the limits -pi/2 and pi/2

and not 0 and pi

[final halo]

draw a picture of the semicircle

[true estuary]

not acc a semi circle

but itll do

[final halo]

right so you want your angle to go from the bottom to the top

anticlockwise

[true estuary]

oh yh

i see it

kl

thx

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@true estuary Has your question been resolved?

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[lean otter]

can you help me with my maths probelm set pls

[safe radish]

Please don't occupy multiple help channels.

[lean otter]

I'm working on this probelm set and I dont have the answers so I ould like to ask you if my answer to this particular question is correct. I got 5cm and 8cm

Here is my working

[safe radish]

@lean otter Has your question been resolved?

[plucky elk]

don't ping random individual people

[lean otter]

I thought he was the helper sorry

[plucky elk]

you ping the helpers role

@helpers

[lean otter]

sorry I'm new here, wont happen again

<@&286206848099549185>

[safe radish]

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[lean otter]

How would I solve this?

$$m \frac{d^{2}h(t)}{dt^2}{} = -G \frac{mM}{(R+h(t))^2}{}$$

[flat frigate]

Brandon H

[hot thistle]

maybe you could start with a change of variables: $\tilde h(t) = R + h(t)$

[flat frigate]

maximo

[hot thistle]

might simplify the problem a little bit

you can also just get rid of the m

so it becomes $\tilde h'' + MG\tilde h^{-2} = 0$

[flat frigate]

maximo

[hot thistle]

this should be separable from here iirc

@lean otter

[lean otter]

I see. Thank you

[hot thistle]

if you need a hint to make it separable || z = tilde h ||

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[swift lily]

so..... how'd you prove the tangent sum identity with euler's formula? I've been stuck on it for a while. All I've managed to do is show that it's equivalent to $\frac{e^{2i(x+y)}-1}{i(e^{2i(x+y)}+1)}$ ... is this even true? I derived this by subbing in sine and cosine with their $e^{ix} - e^{-ix}$ stuff btw

[flat frigate]

Kiameimon | Welt Rene (glomed)

[swift lily]

there's also another trig identity which I'm trying to prove, but it's far harder than this one so I'll stick to this one first

This also got me thinking... usually, we express complex numbers in the form $re^{ix}$ (or rather, we can). But what if $r$, which is meant to be the magnitude, equals to some complex value, like $i$? Do we just "heck care" and just bring it to the power of $e$ with $\ln$?

[flat frigate]

Kiameimon | Welt Rene (glomed)

[plucky elk]

Try multiplying by the complex conjugate of the denominator to the top and bottom

[plucky elk]

r is real

i=e^(i pi /2)

[swift lily]

ah so we just bring raise e to the power of ln(i) in that case

[swift lily]

okie on it

wait, is it necessary to convert it to polar/rect form to find the conjugate? I just realised I never attempted finding conjugates via exp form.

Oh wait.

Just multiply the exponent by -1 the angle to... mb or is it? Was thinking so since conj of complex number maintains same angle but "flips" it around

$\frac{e^{2i(x+y)}-1}{e^{2i(x+y) \ln i}+e^{\frac{\pi}{2}i}}$

[flat frigate]

Kiameimon | Welt Rene (glomed)

[swift lily]

or did I screw it up 💀

(I haven't multiplied by conj yet, just expanding the terms, sorta)

if I did it right then it's equivalent to $\frac{e^{2i(x+y)}+e^{i \pi}}{e^{i(x+y) \pi}+e^{\frac{\pi}{2}i}}$ (idk man at this point I'm just converting everything to exp form)

[flat frigate]

Kiameimon | Welt Rene (glomed)

[swift lily]

I'm honestly just avoiding converting them to rect form like the plague 💀

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@swift lily Has your question been resolved?

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@swift lily Has your question been resolved?

[full girder]

sometimes the very thing you avoid gives you the answer, so if nothing else comes up, try it

[swift lily]

tbh doing so gets me nowhere; I'm not sure how to convert this to rect form

$\cos(2x+2y) + i \sin(2x+2y) - 1$ for the numerator

[flat frigate]

Kiameimon | Welt Rene (glomed)

[swift lily]

... well it's possible I suppose with sum to product? But That won't simplify things

[timid pasture]

by definition tanx is sinx/cosx so tan(a+b) = sin(a+b)/cos(a+b)

[swift lily]

mhm

well it's possible to just prove sin(a+b) = sin(a)cos(b) - sin(b)cos(a) and since cos(a+b) is cos(a)cos(b) - sin(a)sin(b) it'll all just "fall in place"

But Is it doable with euler's formula?

[timid pasture]

yes

use $e^{\theta + \phi} = e^{\theta} e^{\phi}$

[flat frigate]

numbpy

[swift lily]

hm.

I mean this is used to prove cos(a+b) and sin(a+b), but I'm unsure how to do it with tan(a+b) aside from just using the trig identities I just mentioned

[timid pasture]

since euler's identity doesn't directly involve tan or cot, I don't think there'd be a direct way to derive tan identities from euler

[swift lily]

ye...

Okay wait about this, now that I try to derive it... I actually have little clue as to how to derive it using the sine and cosine expansions 💀 it'll result in $\frac{\sin a \cos b}{\cos a \cos b - \sin a \sin b} + \frac{\sin b \cos a}{\cos a \cos b - \sin a \sin b}$, but there's just so many ways to continue from there ig I'll just try spamming other trig identities and see what works

[flat frigate]

Kiameimon | Welt Rene (glomed)

[swift lily]

nani?

[timid pasture]

STOP using emoji with tex

all those weird things are texbot trying to render an emoji

[swift lily]

oh.

[swift lily]

alright, I'm done with that but this... is an absolute monster/

how're you supposed to even prove this?

💀

[jolly trench]

this mildly reminds me of how you can write a fourier series as either a sum of complex exponents or by splitting the imaginary part into a sine and the real part into a cosine

[swift lily]

so like translating between polar and exp form?

[jolly trench]

id imagine you can also just write cos(x+phi) as cos(x)cos(phi)-sin(x)sin(phi) but that seems a bit cheap

[swift lily]

"cheap"

I'll just close the channel, will get back to this sometime in the future

.close

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[pine tulip]

Let the function $f be given: \mathbb{R}^2 \rightarrow \mathbb{R}$, defined by $f(x, y)=\frac{x \sin \left(y^2\right)}{\sqrt{x^2+y^2}}$ if $(x, y) \neq(0,0)$ and $f(0,0)=0$. Study continuity, derivability and differentiability of $f$ in $(0,0)$.

[flat frigate]

Lyuka

[pine tulip]

I already verified the continuity in (0,0). When i try to study its derivability i get this (here i tried to verify it for x)

nvm, i understood my error !

.close

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[short birch]

Hello how do you solve this?

[grim plover]

Let the distance travelled by A at which A and B r together be x. So for B it will be x + (2 * 30) which is x+60 (we do 2*30 since in 2 hrs the extra distance by A will be time * speed)

Now they must cover this distance in the same time, say t

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[lethal root]

explain this

[lean otter]

$$ a = e^{ln (a)} $$

[lethal root]

ooh

[flat frigate]

Brandon H

[lethal root]

ty!

.close

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[obtuse plover]

[lean otter]

[lethal root]

y'all creeping me out

[obtuse plover]

??

[lethal root]

i feel like i'm the only one who does so many mistakes

simple ones

and people are looking at me like

[obtuse plover]

Don’t worry

Ur not

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[cunning tundra]

Quick question when checking for coverergence of a series can i use the alternating series test when (-1)^n and not (-1)^n-1

[junior smelt]

Just a matter of rearranging to remove a negative sign from the series tbh

Regardless you still have something alternating (provided the other “sequence” remains positive!)

[cunning tundra]

Huh explain

[cunning tundra]

Soo yes?

Okay good goood

So this series would be convergent by the alternating series test right

Okay ty ty

❤️❤️

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[supple thunder]

How to solve this

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[supple thunder]

.close

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[lean otter]

$$ f(x,y) = xy +1 $$
$$ \int f(x,y) $$

How?

[flat frigate]

Brandon H

[lean otter]

What is another way to write f(x,y)

in functions with one variable

f(x) can be written as y

how is f(x,y) rewritten?

Should I had started multivariable calculus before differential equations?