help-36

[soft zealot]

knief

[proven lion]

oh o.O

terrifying

[rocky tusk]

now we just isolate dy/dx

can you do that

[proven lion]

he never really taught us how to treat dy/dx as part of the equation that I remember, I just know it as a sign of what to derive :')

and to separate stuff

from ~constants~

[rocky tusk]

just divide by sec^2

(same thing as multiplying by cos^2)

[proven lion]

o

1/sec^2(y)=dy/dx ?

[rocky tusk]

yes which is just cos^2

[proven lion]

fair

[rocky tusk]

so we can now express cos^2 y in terms of x

so if tan(y) = x, can you express this in a triangle

draw a triangle with one angle being y

then the opposite side is x

adjacent side is 1

then find what cos(y) is

and just square it

[proven lion]

x^2?

[rocky tusk]

no thats tan^2(y)

cos is adjacent over hypotenuse

[proven lion]

I did not draw this thing with a hypotenuse :')

[rocky tusk]

hypotenuse is not 1

[proven lion]

waitwhat

[rocky tusk]

use pythagorean theorem for hypotenuse

[proven lion]

OH YEAH

I forgot it's spelled sohcahtoa xD

why am I blanking. This stuff is so easy- sobbb, sec

[rocky tusk]

x^2 + 1^2 = hypotenuse^2

[proven lion]

can I plug in tan(y) for the x ?

waitno

[rocky tusk]

you can but the point was to express cos^2 y in terms of x

[proven lion]

I'm sorry, I'm mostly a visual learner. I appreciate your patience xD

[proven lion]

Isee

[rocky tusk]

because we want dy/dx in terms of x

not y

[proven lion]

ok I'm pitifully stumped

and very tired

how is it almost 7

ooomg I'ma hafta pull another all nighter. I can see it now -.-

[rocky tusk]

happens

caffeine

🙏🏻

[proven lion]

the test is tmr tho, I can't afford this. :')

[rocky tusk]

oh

[proven lion]

mmm I do have some instant coffee

[rocky tusk]

that’s not good

[proven lion]

no, no it is not

it's ok, I got a 48/50 on the midterm and the homework is completion, so I can definitely pass. I've kinda given up on an A tho xD

[rocky tusk]

just take the sqrt

for the hypotenuse

[rocky tusk]

that’s good

[proven lion]

I mean wouldn't that just be x+1=h

[rocky tusk]

no

[proven lion]

well frick

[rocky tusk]

$\sqrt{x^2+1}$

[soft zealot]

knief

[rocky tusk]

$\sqrt{a^2+b^2} \neq a+b$

[soft zealot]

knief

[proven lion]

oh I was doing them separately oops

brb a min, I am being called

by father dearest

baaaack

[rocky tusk]

hello

[proven lion]

helllo

[rocky tusk]

so this is our hypotenuse

[proven lion]

oH

I see what you did

[rocky tusk]

$\cos y = \frac{1}{\sqrt{x^2+1}}$

[soft zealot]

knief

[rocky tusk]

now just square both sides

because we want cos^2

$\cos^2 y = \frac{1}{x^2+1}$

[soft zealot]

knief

[proven lion]

makes sense

[rocky tusk]

now

$\frac{dy}{dx} = \frac{1}{x^2+1}$

[soft zealot]

knief

[proven lion]

ahhhh

[rocky tusk]

problem solved

[proven lion]

ok, I see! :o

need to touch up on my trig :')

I appreciate your help and patience <3

I should probably go speedrun now :<

[rocky tusk]

calculus is only hard if you haven’t mastered algebra/trig

[rocky tusk]

yep

[proven lion]

yeeeeeeeah

[rocky tusk]

good luck

[proven lion]

I thought I was good at algebra, but I think covid kinda whacked me in HS :')

thank you! :)

have a lovely day/night o/

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[rocky tusk]

you too

[proven lion]

<3

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[gray acorn]

lets say i get a linear system like this

[gray acorn]

its 0 solution right? since it contradicts

[velvet yarrow]

what does it contradict?

you mean it's inconsistent?

oh 0x + 0y + 0z = -17.83333?

[gray acorn]

so no solution

right

[velvet yarrow]

yeah no solutiion

[gray acorn]

ok ty

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[drifting harness]

hi i need help with functions and their equations

[warm python]

what's $\Delta$? The change in $y$?

[soft zealot]

ƒ(Why am. I here)=I don't Know

[drifting harness]

i know it

but i dont know the equations

[final saddle]

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[tight lynx]

It gonna be simple question

[tight lynx]

Is a time complexity of O(exponential) classified as an NP problem rather than a P problem?

I have tried to check the difference between definitions of NP problem and P problems, but I am confused due to inconsistency of information.

[tranquil pine]

P problem

[tight lynx]

Why is that?

[tranquil pine]

NP problems are basically problems which you don't know how long it will take to solve deterministically, while P problems can always be solved in a given polynomial time order

[tranquil pine]

Definition

P is polynomial time, NP is non-polynomial time

[tight lynx]

So big O (k^n) is regarded as polynomial.

[tranquil pine]

Yes

[tight lynx]

I didn't know that O(2^n) is even polynomial-

Thank you

Then what are not polynomials for time complexity?

[tranquil pine]

Given a number, find prime factors

I actually forgot why lemme check rq

Since we can always find prime factors in O(sqrt(n))?

[tight lynx]

I appreciate your help.

I get it

[tranquil pine]

im sorry

[tight lynx]

Nah it's fine

I get it.

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[soft cave]

Quick question regarding transitiveness in set theory.
Wouldn't a point being related to itself fulfill the transitiveness definition?

[icy mortar]

Not necessarily.
Transitivity of a relation R on a set A means:if a, R, b and b, R, c, then a, R,c for all a, b, c an element in A.

On the other hand, a point being related to itself refers to reflexivity, where a, R, a for all a in A . Reflexivity by itself does not imply transitivity unless the other conditions of transitivity are met. Does that make sense?

[golden slate]

To add, all 3 conditions for an equivalence relation are completely independent of each other, we can be any combination of reflexive, transitive, and symmetric

[icy mortar]

Exactly

[soft cave]

It's not intuitive to me where the lines are drawn so to say, because an element can be related to itself as part of being transitive? For example (1,1) (1,2)

[icy mortar]

I get it.

transitivity can involve self-relations, but remember that self-relations are neither required nor sufficient to define transitivity on their own.

A relation R on a set A is transitive if, whenever a, R, b and b, R, c, it follows that a, R, c.

Self-relations like (1,1) can participate in the logic of transitivity, but they are not “inherently” what makes a relation transitive. Instead, remember transitivity depends on the chain a, R, b and b, R, c.

[soft cave]

Good explanation, I follow up to this point.
What I don't quite get is, can a and b be the same element?
I believe yes, then can b and c be the same element? I believe no?

[soft cave]

And when you say participate I assume that (1,1) can be part of the logic, it can EITHER be a R b or b R c?

[golden slate]

There’s no restrictions, so I don’t see why not

[soft cave]

Then wouldn't it just be (1,1), (1,1), (1,1) is a transitive relation?

[golden slate]

If 1 R 2 and 2 R 2, then by transitivity 1 R 2, which you already know

[golden slate]

Yeah it’s just a stupid one

[soft cave]

okay one last question

Which one of these two would a (1,1), (1,1), (1,1) relation be?
Is the 2nd case even possible

[icy mortar]

Yes, a and b can absolutely be the same element in the context of transitivity. For example, if (1,1) in R and (1,2) in R, the transitivity condition asks whether (1,2) must also be in R and it already is. So this works fine.

[soft cave]

Actually it must be the 1st one, but then every reflexive relation would be transitive which doesn't make sense

Something is wronnngg

[golden slate]

If you think of a relation as a subset of A x A, then the left is {(1,1)} and the right is {(1,1), (1,1)), (1,1)}, which is an equality of sets

Sets don’t care about duplication

[soft cave]

It's all good gang I'll just tell me professor im not made for this life

Actually my book touched on just this question just after where I was 💀 I think I got my answer, thanks !

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[mortal berry]

but must the eigenvalues ​​of a matrix be arranged in the diagonal matrix in a certain order, or does it not matter how I arrange them?

[eternal topaz]

youre talking about the P and D matrices?

$A=PDP^{-1}$

[soft zealot]

Bonk

[bright pivot]

You are diagonalising, so the order will affect your P matrix

[eternal topaz]

im thinking of smth else

sry

[mortal berry]

[
\begin{bmatrix}
-1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 3
\end{bmatrix}
]

[soft zealot]

Task Bot

[mortal berry]

For example, if the eigenvalues ​​are -1,1,3

For example, I can now exchange the 3 with the -1

[
\begin{bmatrix}
3& 0 & 0 \
0 & 1 & 0 \
0 & 0 & -1
\end{bmatrix}
]

[soft zealot]

Task Bot

[bright pivot]

You can but then you will need to swap the columns in the P matrix

[mortal berry]

And how do I know in what order I should write them

[severe canyon]

In the same order of the respective eigenvalues

So first column of P is the eigenvector associated to the first eigenvalue appearing in the matrix D

And so on

In some context (such as PCA) eigenvalues might be usually written from highest to smallest

[mortal berry]

What I wrote is my D matrix

[severe canyon]

Yes

[mortal berry]

I mean

[
\begin{bmatrix}
2 & 1 & 0 \
1 & 2 & 0 \
0 & 0 & -1
\end{bmatrix}
]

[soft zealot]

Task Bot

[mortal berry]

Of this matrix I found its eigenvalues ​​to be -1,1,3

Now I want to arrange them on the diagonal matrix

[
\begin{bmatrix}
? & 0 & 0 \
0 & ? & 0 \
0 & 0 & ?
\end{bmatrix}
]

[soft zealot]

Task Bot

[bright pivot]

Do u know what diagonalisation achieves

[mortal berry]

But I don't know what order to put them in

[frank ginkgo]

hi

[bright pivot]

@mortal berry

[soft zealot]

jandro0103

[bright pivot]

Is this chatgpt 😭

[frank ginkgo]

i dont understand your question

so i asked an ai

[eternal topaz]

it doesnt matter, as long as the eigenvalue is in the same column as its eigenvector

[eternal topaz]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[frank ginkgo]

is this what you mean?

[frank ginkgo]

its my translator

im not english

[bright pivot]

I think you are really trying at something without actually looking into what it is you are achieving

[severe canyon]

Again, the order doesn't matter

[bright pivot]

Why r u putting Eigenvalues onto a diagonal ?

If u answer what u r achieving with diagonalisation itself, then ur answer becomes very clear

[severe canyon]

Because that's how you construct the diagonal matrix D !?

[mortal berry]

To find the diagonal matrix D

[bright pivot]

I'm asking taskbot

[severe canyon]

Ah sorry

[bright pivot]

Right, what's the point of finding D

Why r u finding D

[mortal berry]

To find the diagonal matrix D

[bright pivot]

Bro, what is the purpose of the diagonal matrix is what I'm asking

[mortal berry]

The text asked me to find the diagonal matrix D

[bright pivot]

Read the text further, the idea is diagonalisation

So the matrix D inherently depends on the matrix P

We are trying to break a matrix A into 3 matrices

[mortal berry]

Now I have to find the diagonalizing matrix P

[bright pivot]

We diagonalise in order to make computations easy

[bright pivot]

So don't you think D will depend on P ?

Or vice versa

[mortal berry]

Yes

Because D=P^-1 AP

[bright pivot]

So D will have eigenvalues, and P will have the corresponding eigenvectors of matrix A

See why diagonalisation works

I think that's a way better way of answering ur question

Because it's a really good question

So it will be rewarding to see for urself

See how and why it is that diagonalisation works

[frank ginkgo]

Hi

do you still need help ?

[bright pivot]

What we are essentially doing in diagonalisation is we are changing the basis from standard basis to the eigenvector basis

And in that basis the matrix operation is a simple stretching

Hence the diagonal matrix

P changes the basis

Diagonal matrix just stretches space

And P inverse gives back the standard basis

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@mortal berry Has your question been resolved?

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[soft zealot]

938c2cc0dcc05f2b68c4287040cfcf71

[final saddle]

@gentle zephyr Has your question been resolved?

[gentle zephyr]

fkk im so tired of this algebra stupid exam

if I manage to pass this exam I will never look back

[forest ember]

dim(W) = 2 -> dim(S) >= 2

So maybe try finding a basis for S then the basis for W must be linearly independent to at least two of the vectors in the basis of S

[gentle zephyr]

dim(S+W) = dim(S)+dim(W) - dim(S ∩ W)

[forest ember]

Is <> span?

If the intersection of S and W is strictly a subset of T then it can it can only be either (4,1,0,2)t or (4,1,2,0)t right?

No nevermind

Because it’s span of T

[tight badger]

is there somone who understand frensh

[final saddle]

@gentle zephyr Has your question been resolved?

[forest ember]

@gentle zephyr
Basis of S, B = {(1,1,0,0), (0,0,1,0), (1,0,0,1)}

One of these vectors, p, will be a basis vector of W, thus S n W = {pt for some t in R | p in <T> }

So p is a linear combination of (4, 1, 0, 2) and (4, 1, 2, 0)

[gentle zephyr]

yeah

[forest ember]

Hmm, none of them can be, so I may have miscomputed B

[gentle zephyr]

S: -x1 + x2 + x4 = 0
S: x1 = x2 + x4
(x1,x2,x3,x4)=(x2+x4,x2,x3,x4)
x2(1,1,0,0)+x3(0,0,1,0)+x4(1,0,0,1)
S = <(1,1,0,0),(0,0,1,0),(1,0,0,1)>

[forest ember]

Yeah

[gentle zephyr]

basis of S is correct sir

we are looking for W
we know
S ∩ W ⊂ T
then
S ∩ T ≠ {0}
S and T have a nontrivial intersection

[forest ember]

By the inclusion-exclusion thm, one v in B must also be in W

[gentle zephyr]

dim(S+W) = dim(S)+dim(W) - dim(S ∩ W)

S + W = R^4
dim(R^4) = dim(S) + 2 - dim(S ∩ W)
4 = 2 + dim(S) - dim(S ∩ W)

[gentle zephyr]

dim(S)=3

[forest ember]

2+3 = 5, 4 = 5 - 1 -> dim(S n W) = 1

[gentle zephyr]

dim(S n W) = 1

[forest ember]

Yeah, we’re missing something

[gentle zephyr]

W = <w1,w2>
w1 ∈ S

w1 ∈ <(1,1,0,0),(0,0,1,0),(1,0,0,1)>

[forest ember]

w1 = a(4,1,0,2) + b(4,1,2,0)

[gentle zephyr]

ye

for a,b ∈ R

[forest ember]

= (4a + 4b, a + b, 2b, 2a)

[gentle zephyr]

yea

[forest ember]

I don’t see it

It definitely can’t be (1,1,0,0 of course

[gentle zephyr]

S ∩ W ⊂ T
S = <(1,1,0,0),(0,0,1,0),(1,0,0,1)>
W = <w1,w2>
T = <(4,1,0,2),(4,1,2,0)>

<(1,1,0,0),(0,0,1,0),(1,0,0,1)> ∩ <w1,w2> ⊂ <(4,1,0,2),(4,1,2,0)>

[forest ember]

(1,0,0,1) = (4a + 4b, a + b, 2b, 2a)
a + b = 0 -> a = - b, 2b = 0 -> b = 0 -> a = 0 which doesn’t work

[gentle zephyr]

this is saying that w1 ∈ T, is it true?

[forest ember]

Yes

[gentle zephyr]

could be a combination of various vectors from S

not necessarily only (1,0,0,1)

lets find SnT

[forest ember]

Ahhh yeah because then it would be l.d.

It only has to be linearly independent to w2

[gentle zephyr]

yeah

T : a(4,1,0,2)+b(4,1,2,0)
T : (4a+4b,a+b,2b,2a)
S : -x1 + x2 + x4 = 0
(x1,x2,x3,x4)=(4a+4b,a+b,2b,2a)
S∩T: -(4a+4b)+(a+b)+2a = 0
SnT: -4a-4b+a+b+2a = 0

a(-4+1+2)+b(-4+1)=0

-a-3b=0

a = -3b

(x1,x2,x3,x4)=(4a+4b,a+b,2b,2a)

(x1,x2,x3,x4)=(4(-3b)+4b,(-3b)+b,2b,2(-3b))

b(-12+4,-3+1,2,-6)

SnT=<(-12+4,-3+1,2,-6)>

SnT=<(-8,-2,2,-6)>

dim(SnT)=1

[forest ember]

(-4, -1, 1, -3) = a(1, 1, 0, 0) + b(0, 0, 1, 0) + c(1, 0, 0, 1) -> a = -1, b = 1, c= -3

Yeah that works

For w1

Now just need w2

[gentle zephyr]

provlem is

we need

S + W = R^4

dim(S) = 3

dim(W) = 2

W = <(-8,-2,2,-6), w2>

S = <(1,1,0,0),(0,0,1,0),(1,0,0,1)>

[forest ember]

Just need a vector in R^4 linearly independent to vectors in basis of S

[gentle zephyr]

yeah

[forest ember]

(0,1,0,0)

[gentle zephyr]

standard basis vector?

[forest ember]

I believe it works

[gentle zephyr]

it works I think

[forest ember]

Another one solved then

[gentle zephyr]

W = <(-8,-2,2,-6), (0,1,0,0)>

[gentle zephyr]

exercises are nasty man

maybe not so much, but sometimes is tricky

[forest ember]

Yeah some of them can take a while to get started

[gentle zephyr]

I appreciate the help

.solved

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[tranquil pine]

2x^2 - 5 = sqrt((x+5)/2)

[tranquil pine]

How would I solve for x

[cyan kayak]

Square both sides, you get a quartic

Quartics suck 🙂

,w solve(2x^2 - 5 = sqrt((x+5)/2), x)

[proper raptor]

i think we need to search a trivial solution

[cyan kayak]

It probably isn't nice, there are techniques you can use to get a closed form solution, but

They're not really practical in most cases

[soft zealot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jan 14, 2025, 6:26 PM

[proper raptor]

And if you know how to do it, you can do a euclidean division of the polynomial $2X^2 -5-sqrt(X+5/2)$ by the polynomial $X-alpha$ where alpha is the trivial solution we found

[soft zealot]

ss

[oblique mica]

notice that you have f(x)=f^-1(x)

[cyan kayak]

Oh! Nice catch

[oblique mica]

all solutions to these problems are f(x)=x

[cyan kayak]

There is a second solution to the problem at around -1.77, but I'm not sure how you would obtain that.

[oblique mica]

,w 2x^2-5=x

[soft zealot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jan 14, 2025, 6:32 PM

[oblique mica]

darn it looks like the bot is down

[cyan kayak]

(1 + √(41))/4 if I mental mathed the quadratic formula correctly

[tranquil pine]

Soo I need to solve a quartic equation?

Are there no other faster methods

[oblique mica]

start here

[tranquil pine]

Oooh

Thanks

.close

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[tranquil pine]

kinda confused how to graph this (precalculus rational functions review)

[tranquil pine]

like if i were to make it into a function

(num with degree 2)/(x+3)(x-3)

[vital crag]

why does it say decreasing on the intervalS when they give one interval

[tranquil pine]

idfk

[vital crag]

did they mean decreasing on each interval separately

[tranquil pine]

presumably

[vital crag]

,w plot (-x^2 + x + 9)/ (x^2-9)

oh that doesn't have horizontal asymptote y=1

[tranquil pine]

that resembles what i got

except with y =

1

[soft zealot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jan 14, 2025, 7:34 PM

[vital crag]

sounds good then

(x^2 + x -9)/ (x^2-9) should fix the horz asymp

[tranquil pine]

yeah thats what i got

okay thank you

.close

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[vital crag]

complain to your teacher about the wording

[tranquil pine]

okay LMFAO

[vital crag]

math teachers love pedantic students

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[fringe quarry]

how the hell do I find points Q and R

[fringe quarry]

<@&286206848099549185>

[scarlet sequoia]

How do you find the equation of a line perpendicular to another

like

[fringe quarry]

I done part a

[scarlet sequoia]

ok

[fringe quarry]

I need to do part b

[scarlet sequoia]

so you have shown QS has equation x+3y = 22

[fringe quarry]

yes

[scarlet sequoia]

well

you know the x coordinate of Q

don't you?

[fringe quarry]

…

should I!

[scarlet sequoia]

the y coordinate

sorry

yes you should

[fringe quarry]

oh u meant y

yes it’s 0

[scarlet sequoia]

ok

so

what's its x coordinate

if y = 0

knowing it's on the line QS...

[fringe quarry]

22

AND R IS 2 ALONG SO ITS (24,6)

I get it now

thank you

bit of a brain fart

but tysm 🙏🏻

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[rustic finch]

could someone possibly double check my work?

[dapper hull]

What were you doing here ?

[final saddle]

@rustic finch Has your question been resolved?

[rustic finch]

compound angle formula

[dapper hull]

i get you but why is it a split fraction
it should be 1 big fraction

[rustic finch]

oh, that's a mistake

is the answer correct though?

[dapper hull]

no

[rustic finch]

so, can you show me how to do it?

[dapper hull]

ok

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[dapper hull]

just reopen the channel

[rustic finch]

.reopen

[final saddle]

✅

[dapper hull]

Here

[rustic finch]

thank you

I understand where I went wrong now

[dapper hull]

don't forget that you should multiply by the conjugate

(when writing the final answer)

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[wanton trout]

Someone help me with this I really do not understand I am so lost…my teacher doesn’t know how to explain it to me properly

[mint orbit]

you might be interested to know that $\int _a ^b f(x) \dd x = \int _a ^c f(x) \dd x + \int _c ^b f(x) \dd x$ if $a \leq b \leq c$ and other assumptions

[soft zealot]

jan Niku

[mint orbit]

can you think of how you might use this fact

[unique totem]

can anyone help me with this question?

[wanton trout]

wait im thinking

[unique totem]

i know the answer is (B) but i do not understand why

[glossy zephyr]

go make your own help post please

[unique totem]

i did

[rapid sky]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[rocky tusk]

this is not your channel bruh

[unique totem]

how do i make my own channel

[rapid sky]

Go to that channel in the message.

[glossy zephyr]

https://discord.com/channels/268882317391429632/488120190538743810

[wanton trout]

um do i add the integral from 0 to 10 and then add the one from 3 to 10??

idk im very slow im so sorry

[unique totem]

okay peace

bye guys

[wanton trout]

oh

would integral of 0 to 3

for f(x) be 19

and then g(x) is 16

so u minus? and then get 3...?

[mint orbit]

we should prolly make a better notation

maybe

or a better way of writing this

[mint orbit]

i think the idea is that, if we want $$\int _0 ^3 h(x) \dd$$, and we know $\int _0 ^{10} h$ and $\int _3^{10}h$ then we might think $$\int _0 ^3 h = \int _0^{10} h - \int _3 ^{10} h$$

[soft zealot]

jan Niku

[mint orbit]

if youre confused about the last line, try adding $\int _3^{10} h$ to both sides, and then using this rule to figure they really are equal

[soft zealot]

jan Niku

[wanton trout]

okay i kind of understand

that is what i did to find f(x) froom 0 to 3

i do the same with g(x)?

[mint orbit]

you also need to use linearity

lets write out the facts we have here ...

[soft zealot]

jan Niku

\begin{enumerate}
\item $\int _0 ^{10} f = 21$ \\
\item $\frac 12 \int _0 ^{10} g = 8$ \\
\item $\int _3 ^{10} (f-g) = 2$
\end{enumerate}

jan Niku

[mint orbit]

@wanton trout can you find the value of $\int _0 ^{10} (f-g)$ using facts 1-3?

[soft zealot]

jan Niku

[wanton trout]

yeah

[mint orbit]

what do you get?

[wanton trout]

its 13

becuse 21-8

[mint orbit]

youre close but youre missing one thing

[wanton trout]

the 1/2?

[mint orbit]

in item 2, its half

yea

[wanton trout]

so 17?

[mint orbit]

hmm?

[wanton trout]

or do i have to half the 13

[mint orbit]

so lets look at 2

$\frac 12 \int g = 8$ then $2 \qty( \frac 12 \int g ) = 2 \qty( 8 )$

[soft zealot]

jan Niku

[mint orbit]

just multiplying by 2 on each side

hows it work out

[wanton trout]

why do we do 2 times 1/2 of g?

[mint orbit]

because we want to know $\int g$ not $\frac 12 \int g$

[soft zealot]

jan Niku

[wanton trout]

ohhhhhhhh

so g(x) would be 16

[mint orbit]

yea, to be more specific

[mint orbit]

well, lets not clutter up the chat too much math, can you just fix your answer here?

[wanton trout]

21-16 = 5

[mint orbit]

yea

[wanton trout]

so 5 - 2 = 3

[soft zealot]

jan Niku

[mint orbit]

yea i think you got the idea

[wanton trout]

if i had something else would it be the same thing?

[mint orbit]

hmm? something else?

[wanton trout]

one sec i take pic

Same idea here right?

[mint orbit]

yea, but you need one more property

[soft zealot]

jan Niku

[wanton trout]

ohh okkk

can you help me with um another question

or do i have to make a new channel

[mint orbit]

i am gonna eat

you should make a new channel

more people will see it

[wanton trout]

okay thank you for help

.close

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[jagged flare]

let there be a non-convex quadilateral ABCD, with angle B=C=60 and angle A=30, line AB and and CD intersect at E, AD and BC intersect at F, and EF and BD intersect at P. If CF=AE=1, find EP^2

[jagged flare]

yea so basically AFB is 90 cause 30 60 90

BEC=60

[full flint]

∆BEC is a equilateral triangle because of b= 60° c= 60°

[jagged flare]

yea

[full flint]

So EA= ED

BA is a straight line

[jagged flare]

oh yeah

[full flint]

BEC is 60

[jagged flare]

so EC=3

cause CDF 30 60 90

[full flint]

mb

[jagged flare]

hmm coordinate bash seems possible

but is there a non bash way?

[full flint]

Let me try to work it out

[jagged flare]

damn this is ugly did i do anything wrong

wait hold on what if mass of points

so EP=3/7 EF

wait im a dumbass ptolemy doesetnt work here lmao

EF is sqrt7

so EP=3/sqrt7?

[fierce orchid]

yes

[jagged flare]

ok ty

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[waxen jay]

How are these two equivalent?

[waxen jay]

And how are you able to get rid of 1/(n+1000) after n = 1000?

[ivory vessel]

index shifting

let u = n + 1000, then the lower bound becomes 1001

[waxen jay]

i'm still a little confused, once you say u = n + 1000, you can replace the demoninator so that it is 1/u

but how does that affeect the index

[ivory vessel]

the sum is summing from n = 1 to infty, right?

[waxen jay]

yes

[ivory vessel]

let u = n + 1000, since n = 1, the lower bound is now u = 1 + 1000 = 1001

the upper bound stays the same, because infty + 1000 = infty

[waxen jay]

oh i see

ok that makes sense now

thank you ‼️

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[sour seal]

i have a final exam tomorrow and i have 0 clue how to do this pls help

[final saddle]

@sour seal Has your question been resolved?

[sour seal]

<@&286206848099549185>

bro no help 💔

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[azure blaze]

1e+308K ×3 =

<@&286206848099549185>

Click 🎉 button to enter!
Winners: 1
Ends: <t:1736961168:R> (Timer)

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[next thorn]

i dont understand how the area of the side triangle in a pyramid is 1/2 * height (thats perpendicular to the edge of the base) * perimeter of the sqaure (the base of the pyramid)

[next thorn]

soo the law for a triangle's area is 1/2 * base * height

[next thorn]

i understand the height in this

[next thorn]

but how is the 1/2 base of the trinagle equal to 1/2 base of the square

[steel plaza]

I might be misunderstanding you but the side triangle has the base which is equal to the side length of the square present below in the pyramid

[next thorn]

yeah i mean, they are equal so

height * 1/2 base = 1/2 height * perimeter of the square

so base of triangle = perimeter of square

[steel plaza]

It shouldn't be perimeter

[next thorn]

that is a war crime

[next thorn]

oh wait

[next thorn]

ahaaaaaa

so the base of the triangle is equal to one of the lengths

aka 1/4 of the perimeter

[steel plaza]

Yes exactly

[next thorn]

ok then wtf is this

[steel plaza]

I wonder the same

[next thorn]

i cant wrap my head around it tbh

the law is 1/2 * perimeter of square * the perpendicular height

that means that the perimeter of the square is equal to the base of the triagle... ist somethings wrong

[azure jacinth]

Its bc the area of the triangle is (1/2)(base)(height) but we know that the base is 1/4 of the perimeter of the square below so we can say (1/2)(1/4 perimeter of square)(height) but they there are 4 triangles so (1/2)(perimeter of the square)(height)

[next thorn]

the LAW is not wrong, this one

so what am i doing wrong

[steel plaza]

That is for 4 triangles I believe

[next thorn]

OOOOH

THE RULE IS LIKE

THE VOLUME OF THE PYRAMID

NOT THE AREA FOR THE SIDE TRIANGLE

💀 mb

[steel plaza]

You sure?

[next thorn]

oh wait

no

fu-

[steel plaza]

Volume of a pyramid's formula looks quite different

[next thorn]

holup

[steel plaza]

The law you stated is actually the lateral surface area of a square pyramid

Meaning area of all the triangles present

[next thorn]

oh so we divided by 4

[steel plaza]

If square has side a and slant height is l

for one triangle it's 1/2 × a × l

for four triangle it's 2×a×l

Your law states it's 1/2 × perimeter × slant height

1/2 × 4a × l

= 2 × a × l

[next thorn]

okay i think i get it now, you're saying my problem is that i thought the law was for ONE triangle, not FOUR

[steel plaza]

Same thing

[steel plaza]

Yes

[next thorn]

yeaaaah

makes a ton of sense now

much appreciated sorry for me being

slow

uhm my finales are tomorrow 😊 bye

[steel plaza]

nah man it's all chill

[steel plaza]

all the best

[next thorn]

.close

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[grim imp]

Idk what to do about the 1/3 exponent

[final tangle]

same derivative laws apply

1/3 is just a number

[grim imp]

Plz explain

Do I like ignore it

[final tangle]

no

[grim imp]

And just do quotient thing

Or do I like apply it to all of the g

[final tangle]

start with quotient law, then continue from there

[grim imp]

do i do it normally or do I apply 1/3 to all the g

[final tangle]

same thing

that's what you'd do normally

[azure jacinth]

Just use quotient rule

[grim imp]

OK do i divide by g^2/3

[azure jacinth]

Yep

Then just substitute those from the table where x = 0

[grim imp]

ok

[final tangle]

you'll have g^(2/3) in the denom after the initial application if that's what you mean

[grim imp]

Im at this idk if it's right so far

[final tangle]

no

don't sub in anything yet

start with just the quotient rule

get that sorted before attempting to evaluate

[azure jacinth]

Dx ( cube rt g) = 1/3 cube root g² not g'

[azure jacinth]

Yea do this first dont plug any values yet

So u dont get confused

[grim imp]

IDK

[final tangle]

do you know the quotient law?

[grim imp]

yes

The bottom is like g^2

So like

1/3 times 2

[scarlet notch]

can someone teach math in german here ?

im new

please my exam is tomorrow

[final tangle]

let's change the g in quotient law to h, so we don't mix it up with the g in your question

(f/h)' = ?

your denominator seemed ok
what would the numerator be when applying that law?

[grim imp]

f'h-fh`

[azure jacinth]

Open a new help channel

[final tangle]

and what is h'

or (g^(1/3))'

[grim imp]

soo

I do apply 1/3 to all of them

[final tangle]

wdym

[grim imp]

like all the g have 1/3 on them

[final tangle]

your denominator is g^(1/3)

that's what you use in your work

[grim imp]

sooo yes

OK I GOT THE ANSWER

It's 7

OK

[grim imp]

This was all I needed to know Fr

Ty

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[golden coyote]

hi yeah so i dont understand e is not a function right

so

i dont get thi

im learning calculus for the first time and i dont understand if this is a composite function or not

because e is a constant no

[zealous storm]

e is not

but e^x is

just like 3 isn’t a function

[golden coyote]

and what would the inner and outer function be

so i understand that

e^x is one

is the outer

-x^2

[zealous storm]

well technically it is

just a straight horizontal line

but yeah

[zealous storm]

yes it’s the composition of e^x and -x^2

[golden coyote]

okay so chain rule would be

-2x * dy/dx e^x right

im sorry im stupid 😔

[zealous storm]

huh did you express it as a composition first

[golden coyote]

i dont really know how to lmao

its -x^2 and e^x right

[zealous storm]

apply this

f(g(x)) is a composition where you pass as input g(x) to f

[golden coyote]

w(u(x)) w(x) = -x^2 u(x) = e^x

or am i wrong

[zealous storm]

so like f(x) = 3x and g(x) = x^2, for example

what’s f(g(x))

[golden coyote]

no flip w(x) and u(x)

[golden coyote]

3x^2 right

[zealous storm]

yes

[zealous storm]

now compute w(u(x)) here

[golden coyote]

w(x) = e^x

u(x) = -x^2

w(u(x)) = w(-x^2) => w(u(x)) = e^(-x^2)

wrong or right

[zealous storm]

good

[golden coyote]

yesss thank you so much

so wait

now lastly

differentiate

i know how to differentiate

-x^2

thats -2x

but i dont know how to differentiate e^x

is it just

xe^(x-1)

[zealous storm]

nah

derivative of 3^x, for example, isn’t x3^(x-1)

[golden coyote]

so how shall i differentiate this

[zealous storm]

have u learnt the derivative of e^x

or just a^x? for appropriate a

[golden coyote]

not yet im new to calculus

so far

i have learnt normal differentiation so i can like differentiate like a^x

but i dont know the rest

[zealous storm]

this is a big deal in the calculus storyline lol

[zealous storm]

okay then same thing

[golden coyote]

my friend said to learn chain rule first

[zealous storm]

how do you differentiate a^x

[golden coyote]

xa^(x-1) no

[zealous storm]

it is chain rule but if u don’t know how to differentiate e^x itself then whats the point

[zealous storm]

no

a^x has derivative a^x ln(a)

what you’re talking about is the power rule or something

[golden coyote]

when i say a i mean just any like real number

[golden coyote]

yeah that

[zealous storm]

the derivative of x^2 is 2x

[golden coyote]

yeah

[zealous storm]

but the derivative of 2^x

is not x 2^(x-1)

[zealous storm]

yeah i’m aware but that’s not how it works

[golden coyote]

so what do you think i should learn after the power rule cos im kind of lost

[zealous storm]

anyway i have to go but the derivative of e^x is just e^x

because ln(e) =1

[zealous storm]

learn the derivative of e^x

u can google it

or just “exponential” or whatever u call it

https://youtu.be/yg_497u6JnA?si=PeEBOxwR_RdVWqFw

watch this

your question is very similar to the exercises in that video

u see there’s even an e^(x^2) there

[final saddle]

@golden coyote Has your question been resolved?

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[velvet frigate]

The base of a right prism is a right triangle with a hypotenuse of 8 cm and an angle of 30 deg
The volume of the prism is 48sqrt(3) cm³. Find the lateral surface area of ​​the prism.

[velvet frigate]

I got 72+24sqrt3

is this correct

[rugged merlin]

Yep

Sounds good to me

[velvet frigate]

aito ty

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[velvet frigate]

.reopen

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[velvet frigate]

The function F(x) = 2x ^ 3 - 1 is an antiderivative of the function f(x) Specify the function f(x)

i believe the answer here is the last one
f(x)=6x^2

is this correct?

[rugged merlin]

Indeed

You're correct again

[velvet frigate]

ngl i have no idea how i solved it, i just found the dash of f(x)

F'(x)

[rugged merlin]

Yep

That is how you're meant to solve it

F'(x) = f(x) by the definition of antiderivative

[velvet frigate]

alright

then I've got another question one sec

Which of the following functions is an antiderivative of the function f(x) = x^-4?

[rugged merlin]

You gotta integrate

[velvet frigate]

the thing is we weren't really taught how to
so in that case I'd have to take each F(x) and take the dash of them?

[rugged merlin]

Ok so you don't know anything about integration

in that case, i guess that is the best approach

one way to think about is like this

[oak plume]

do (x^n+1)/n+1

[rugged merlin]

"what function, if i differentiate it, gets me x^(-4)?"

[velvet frigate]

differentiate it being the dash thing?

[rugged merlin]

yea

do x^(-4) is the result of differentiating another function you dont know about

or doing the "dash thing"

[velvet frigate]

yea sorry i study in my native language I didn't know it's differentiating)

[rugged merlin]

so give it a think

what function, if you differentiate it, gives u x^(-4)?

remember that if f(x) = x^n then f'(x) = nx^(n-1)

[velvet frigate]

theres a seperate formula for fractions right

[rugged merlin]

"fractions"?

can you specify

[velvet frigate]

one sec

[rugged merlin]

oh it is called the quotient rule

no don't worry about that here

it is not relevent

this is what u need

[velvet frigate]

so do i just need to find the dash of (x^-4)

[rugged merlin]

No, you're misunderstanding

[velvet frigate]

a would be -3

[soft zealot]

Aero

[rugged merlin]

yes, thats almost right!

but issue is that if you take it as F(x) = x^(-3), then f(x) = -3x^(-4)

NOT f(x) = x^(-4), which is what u have

how do we get rid of the -3?

[velvet frigate]

f(x)

erm

idk im a bit confused about that part

[rugged merlin]

so, do you understand that this is what you're trying to do to be clear?

[velvet frigate]

yea

[rugged merlin]

so you were correct that the exponent ??? is meant to be -3

[soft zealot]

Aero

[rugged merlin]

what is the remaining ??? meant to be though

[velvet frigate]

wait is it not the same thing

its both meant to be a

[rugged merlin]

so you are saying it should be -3x^(-3), yes?

[velvet frigate]

yea

[rugged merlin]

ok so

try to differentiate that

what is (-3x^(-3))'?

[velvet frigate]

-3*

i dont remember ngl

-3 times something

and x^-4 i think

[rugged merlin]

well apply this

[velvet frigate]

ok so -3*-3^-4

[rugged merlin]

you missed the x

[velvet frigate]

-3*-3x^-4

[rugged merlin]

yea

so

is that x^(-4)?

[velvet frigate]

it's 9x^-4 right?

[rugged merlin]

yes

[velvet frigate]

ok so no

[rugged merlin]

issue is you want it to be x^(-4)

try the opposite

you are trying to get rid of the -3

how do you cancel out a -3?

[velvet frigate]

i need to rid of which -3 again?)

[velvet frigate]

here?

[rugged merlin]

so you see how when you differentiate it it brings down a -3?

you want to get rid of it to get only x^(-4)

[velvet frigate]

or well just divide both sides by -3
we have an equation right 😅

[rugged merlin]

you are very close!

[soft zealot]

Aero

[rugged merlin]

the question is what must ??? be here

[velvet frigate]

gotcha

i know i'd divide it by -3
and -3 is also 1/3^-1

or no

what

[rugged merlin]

-3 is not the smae 1/3^(-1)

[velvet frigate]

-3 is -1/3

[rugged merlin]

i mean, that statement doesn't make sense

do you mean ??? is -1/3?

[velvet frigate]

yes

[rugged merlin]

then yeah

it works out

because -1/3 * (-3) = 1

[velvet frigate]

but don't both things have to be same

like on the left

[rugged merlin]

they do not

maybe i should recolour it

[velvet frigate]

theres x^-3

oh ok

[soft zealot]

Aero

[velvet frigate]

Is there any way you'd be able to go over this in voice chat? If not I understand it's just I really want to recap everything we did

[rugged merlin]

i dont think this server has public voice chats

unless I can't see them

[velvet frigate]

Yes the server doesn't have them but dms should work
If you're not up to that totally get it

[final saddle]

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[ivory bough]

Can someone help me understand why hermitian matrices can always be diagonalized with unitary matrices? I know all eigenvalues of hermitian are real and eigenvectors are orthogonal but I am not sure how to use that info in this case

[tiny gorge]

this is (one form of) the spectral theorem

have you proved that?

if not then you have most of the pieces you need

eigenvectors corresponding to distinct eigenvalues will be automatically orthogonal

and for any eigenvalues that have geometric multiplicity > 1 you can take any basis for the eigenspace and orthogonalize it using gram schmidt

so the remaining piece is to show that the eigenspaces have full geometric multiplicity (equal to algebraic multiplicity)

[ivory bough]

No

[ivory bough]

is gram schmidt like u = (A-lambaI)w where w is known non eigenvector so u(A-lambaI) = 0 as (A-lambaI)(A-lambaI) = 0

thus u is eigenvector to A?

[tiny gorge]

gram schmidt is just a simple algorithm for making an orthonormal basis given an arbitrary basis

[ivory bough]

oh ok

thanks

[tiny gorge]

the hard part of the proof is showing that a basis of eigenvectors exists

for that you should read a proof of the spectral theorem, one way is to show that if A is hermitian (or more generally, normal) then A and A* have a common eigenvector

and then you use induction by looking at the space orthogonal to that eigenvector

[ivory bough]

thx

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[knotty night]

I got one simple question. How come the restrictions include positive 2?

[knotty night]

Do I factor first and then do the reciprocal?

Real simple

[drowsy epoch]

You are starting off with a fraction where you divide by x²+x-6 which gets 0 if x = 2

[knotty night]

you mean factor?

oh

Yeah, I see it

I usualy just factor and find the restrictions through that

So the restrictions is any value that when subbed into X, will make 0 in the denominator while simplifying?

@drowsy epoch

[drowsy epoch]

yes

never forget from where you came from

if this is your starting point, then find first the domain from here

[knotty night]

alright, thanks

[drowsy epoch]

and as you progress you might find more restrictions

[warm ether]

(dont mind the bunny)

[buoyant solar]

hey i need help

[drowsy epoch]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[buoyant solar]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[knotty night]

#help-40

[final saddle]

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[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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[fiery hedge]

Hello I need help with a question, I suppose I hsould just post it here?

[fiery hedge]

suppose I have a degree 4 polynomial in say 10 variables, and I know that for its positive semi-definite for all real values, so its at least \geq 0.

The question is: is there a normal form for the polynomial? For example, does it mean f = f1^2 + f2^2 + … where fi are all degree two polynomials?

For example, suppose f is degree 2, then this follows from the diagonalization of bilinear pairings.

maybe we should just consider the say in 2 variables

if we know that ax^4 + b x y^3 + c x^2 y ^2 + d x^3 y + e y^4 is positive semi-definite

okay appearantly this is Hilbert's 17th problem LOL

and appearantly this is the answer

[final saddle]

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[prime forge]

Would anyone be able to help me remember why I've done this? (Or rather why I copied it from the blackboard) How did we get rid of arctan(1-x)?

[winter lava]

can we say lim x->0 f(x) = lim x->0 f(-x)?

and apply that to the second limit

[prime forge]

Yes? Though I still don't see it

It is true for arctan, but that wouldn't that leave us with (pi/4 + arctg(x-1))/-x?

No, sorry

it would be (x+1)

[winter lava]

$\lim_{x\to 0} \frac{\frac{\pi}{4}-\arctan(1-x)}{x}=\lim_{x\to 0} \frac{\frac{\pi}{4}-\arctan(1-(-x))}{-x}$

[soft zealot]

Axe

[prime forge]

awesome, thanks

.close

[final saddle]

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[bleak edge]

is this the integral symbol

[craggy bloom]

Could be but i dont know

[silver basin]

I don't think so. Probably the one for like 3/2 but written as 1 1/2

[slender cargo]

no

[raw marten]

is this channel still occupied?

[silver basin]

nah

[hoary saffron]

May ABCDE be an unregulated pentagon, and M,N,P,Q,U,V midpoints of AB AE ED BC MP NQ in this order. If UVvectotial = a* DC vectorial, find a(a real number) 9th gradr problem

[copper roost]

im kind of confused

so the period of the sinus function is 1