help-49

[waxen silo]

cosh/cosh is just u :V

[jagged saffron]

Well its 1

[waxen silo]

yea, i intergrated already, my mistake

[jagged saffron]

Right, good!

[waxen silo]

alright, thanks for the help!

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[misty marlin]

Hi. Should i simplify it further or?

[midnight plank]

@misty marlin Has your question been resolved?

[misty marlin]

<@&286206848099549185>

[wraith orchid]

yehee

[misty marlin]

I should?

[misty marlin]

this sounded like Michael Jackson in my head

[wraith orchid]

ook

what''s your question???

[misty marlin]

.

[wraith orchid]

there's a problem in soln

[misty marlin]

Mistake?

[wraith orchid]

2nd last line

yes mistake

you just can't terminate numerator values when they are in addition or substraction format

[misty marlin]

I see that was dumb mistake

[wraith orchid]

not a problem.... just solve that and then see...

[misty marlin]

Okay

[misty marlin]

Should i simplify it further or no

[wraith orchid]

nope.....

that's it ig

[misty marlin]

what bout cos4t

[wraith orchid]

or you can try further

[last slate]

Simplify it further

[worthy wing]

?

[misty marlin]

Sure

[last slate]

The ans will come cotx

[wraith orchid]

wait you taking it that way........

[misty marlin]

Ans?

[worthy wing]

You recheck your work

[misty marlin]

Cotx is ctgx right

[wraith orchid]

well just write the question for me

[misty marlin]

Simplify

[last slate]

Yep

[wraith orchid]

the trigonometric question

[misty marlin]

Sure thing

Thanks

[last slate]

Wlcm ig ^^

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[misty marlin]

Thank you. I just got the same.

[last slate]

Wlcm ✌️

[misty marlin]

Thank y'all

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[grizzled torrent]

If the formula in the graph of the first picture (PMF) represents the probability of getting EXACTLY k successes after n trials (with a fixed success rate = p) and the one in the second pic represents the probability to get AT LEAST k success (CDF), what is the formula to determine the probability of getting AT MOST k successes after n trials? Is it just 1-CDF?

[grizzled torrent]

I'm confused because the respective functions on python look like this, instead of the other way around:

def calculate_probability_at_least_k(k, n, p):
    probability = 1 - binom.cdf(k - 1, n, p)
    return probability

def calculate_probability_at_most_k(k, n, p):
    probability = binom.cdf(k, n, p)
    return probability

[midnight plank]

@grizzled torrent Has your question been resolved?

[midnight plank]

@grizzled torrent Has your question been resolved?

[midnight plank]

@grizzled torrent Has your question been resolved?

[midnight plank]

@grizzled torrent Has your question been resolved?

[lilac nebula]

Yeah usually the CDF is a monotonically increasing function and displays the probability for k or less than k successes

That's why it's cumulative, b/c if you go from left to right (i.e. from k= -infty to k= +infty) you add up all the previous values of the PDF essentially

So I'd say the screenshot has the wrong idea when it just writes "CDF" for this situation

What the screenshot display is, to me, the function 1 - CDF where CDF is the cumulative distribution function associated to the PDF from the lower graph in the first picture

[grizzled torrent]

okay, but are they both (at least, at most) CDFs? btw, During this time I realized that the function I was looking for is not in fact 1 - CDF

can you confirm for me that these are the right functions and the graphs look alright?
at least k:

at most k:

[lilac nebula]

By definition a CDF associated to a probability distribution is the function that associates a number k with the probabilty of getting an outcome smaller or equal to that number

[grizzled torrent]

ah okay, so the upper ones aren't CDF

[lilac nebula]

Yea I wouldn't say so

[grizzled torrent]

would they have a name?

[lilac nebula]

I mean the first screenshot is also something cumulative, but it doesn't come from a probability distribution in the same way

It's just 1 - CDF, I don#t think that has a name

The screenshots look alright

[grizzled torrent]

I don't think it is 1-CDF

for example, let me think

n=4, k=4, p=1/2
P(X ≥ k) = 1/2⁴
P(X ≤ k) = 1

no wait that's not the point

I think you're right then

[grizzled torrent]

so, in this case, if the bottom graph is the CDF(k;n,p), the top one is some f(n;k,p) equal to 1-CDF(k;n,p)?

[lilac nebula]

yeah sounds alright

[grizzled torrent]

thank you for the help

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[bleak pier]

Hints please

[fresh sparrow]

can you find the formula for the nth term of the series?

[bleak pier]

Yes

[misty gorge]

if you're trying to be cheeky and use the answer choices, there are two of these that are already way too small to be the right answer

[bleak pier]

Multiplication (1 to n) (2n-1)/4^n

[bleak pier]

You meant by adding first few terms

@wary thorn

[misty gorge]

in fact, the first term is enough to make you eliminate one of them

[wary thorn]

I've sent you it already

[bleak pier]

Yes

B and d

Multiplcation of 4

Any multiplication sinx/x formula?

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@bleak pier Has your question been resolved?

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@bleak pier Has your question been resolved?

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[mystic salmon]

I need to find the inverse of the function f(x) = x^2 + 10x + 28. This is what I've got so far, but I am confused as to how I would continue:
y = x^2 + 10x + 28
x = y^2 + 10y + 28
x - 28 = y^10 + 10y

I would factor out the y, but I don't want to define f(x) in terms of y.

[distant heron]

The first equation is sufficient, you need to solve in terms of x

[mystic salmon]

I don't get what you mean by that. What's the 'first equation' you're referring to?

[distant heron]

The first equation of the list

y = x^2 + 10x + 28

This becomes x^2 + 10x + 28-y = 0

If you consider y fixed it becomes a polynomial equation you can solve in terms of x

[mystic salmon]

ohhh

ok thanks

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[winter imp]

Let k be the circle described around the triangle ABC. If P and Q are the points where the bisector BC intersects the lines AB and AC, prove that Q is the inversion of P. My friend who is grounded needs help and he cannot use discord.

[winter imp]

Wait he managed to do it

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[tepid rune]

is my mistake the fact that i turned the hyperbolic trig functions into their complex forms?

[last slate]

hi

[tepid rune]

my issue is when i get to another part of the question where i'm supposed to find max curvature, there isn't a t i can work with

<@&286206848099549185>

nevermind i got it

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[wispy coyote]

can someone explain? is this some log property?

[wispy coyote]

honestly dont even know what it says

x/16^x -1?

[flat veldt]

What is the question

[burnt flame]

$\lim_{n \to 0} \f{25^{n}-1}{16^{n}-1}$?

wait

is it 0?

[grand pond]

🫎 Chmoosey 🫎

[flat veldt]

Lhopital

[wispy coyote]

x instead of n i think but yeah

[burnt flame]

they appear to be multiplying numerator and denominator by x?

[wispy coyote]

yes limit approaches 0

[burnt flame]

is this your work?

are you just confused on how to do it?

[wispy coyote]

my internet.....
somehow it says
25^x - 1 / 16^x - 1 = (25^x - 1 / x * x/16^x-1)

[wispy coyote]

my friend's trying to practice for exam on webnesday

[wispy coyote]

iam confused how it equals that

[burnt flame]

they just appear to be multiplying numerator and denominator by x, they are multiplying by 1 essentially

[flat veldt]

The 2 xs cancel each other out

[wispy coyote]

oh

i didnt know you could do that....

aight, thanks

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[radiant vector]

A bit stick in terms of how to create an equation for this

I thought abt creating an equation in terms of the difference (the difference differs each by 1)

so 2(n-1)1 but idk

[vivid whale]

WHERE IS DR SLEEP?

i demand an audience

[radiant vector]

hm?

[fathom tendon]

Do you know the formula for the sum of
1 + 2 + 3 + ... + n

[radiant vector]

Yup

[fathom tendon]

What is it

[radiant vector]

n/2(a+l) l = last value that is if you have the last value if not then the second equation is

n/2 ((2a+(n-1)d)

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[cunning zodiac]

ax+by=10
2x-ay=0

solve the pair of simultaneous equations for x and y, expressing your answer in terms of a and b

i keep on getting 0 = 0 somehow

[sage olive]

what did you try?

[cunning zodiac]

making x the subject

ad subbing into other equation

[sage olive]

what did you get for x

when making it the subject

[cunning zodiac]

ay/2

x=ay/2

[sage olive]

yeah if 2x - ay = 0 then x = ay/2, then you substituted that into the first equation?

[cunning zodiac]

ye

[sage olive]

what did you get when substituting

[cunning zodiac]

((a^2*y)/2)+by=10 ?

[sage olive]

yeah

[grand pond]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[sage olive]

yeah that looks right

so what did you get when you solved for y?

[cunning zodiac]

like made it the subject?

y=2x/a

[sage olive]

no when you solved this equation

the whole point of the substitution was that you turned an equation that had both x and y into an equation that just has y

then you solve for y

and then lastly you plug that back into your substitution to solve for x

[cunning zodiac]

hm its quite hard to type

nvm i got it

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[vague burrow]

i want the determinant of those questions and whatever they say too

[merry pewter]

what

[vague burrow]

idk how to solve these 2

[merry pewter]

for the first one, did you compute AB?

[vague burrow]

i have multiplied AB

idk how to do the eqution thing

equation*

[merry pewter]

what do you get

[vague burrow]

a matrix

[merry pewter]

sick

what is the matrix

[vague burrow]

i dont hace a phone to send the pic rn:((

[merry pewter]

can you type it

[vague burrow]

i just wanna know what to do w the equations

[merry pewter]

what you get if you compute AB is probably important to find that

i was thinking that if it was the identity times a constant, then you would have found the inverse up to a multiplying constant

[vague burrow]

i think

the matrix i got

i have to multiply it w x y z

and keep it equal to 10 8 and 7

yes i got it lmao

[vague burrow]

can u help w this one? i got no shittin idea what to do

oh wait

[merry pewter]

this is what i was saying earlier, you find the inverse, and multiply on both sides to solve the equation

,, Ax = b, x=A^{-1}b

[vague burrow]

okay nvm i didnt get it

[grand pond]

Bair

[vague burrow]

can u maybe type it or write it:((?

[merry pewter]

write what

[vague burrow]

like how to do the second one

i didnt get it

i understand the formula idk how to apply it

[merry pewter]

you are given A and b, and are asked to solve for x

[vague burrow]

i get that

no im just given a matrix A

i gotta fiind the a^-1

[last slate]

kya quality hai 🤣🤣

mc quality

[merry pewter]

you are given that
$$\begin{bmatrix} 1 & 2 & -3 \ 2 & 0 & -3 \ 1 & 2 & 0 \ \end{bmatrix} \begin{pmatrix} x \ y \ z\end{pmatrix} =\begin{pmatrix}1 \ 2 \ 3\end{pmatrix}$$

[grand pond]

Bair

[last slate]

left hand multiply both sides by the inverse of the coefficent matrix

[merry pewter]

yeah thanks bro

[last slate]

$$ Ax = b, x=A^{-1}b
A^-1 A x = A^-1 b
I x = A^-1 b
x = A^-1 b $$

[grand pond]

kr

[last slate]

oh bruh

im mid as hell at latex

@merry pewter some help bro 🤣

[merry pewter]

i already wrote it

[last slate]

oh

[merry pewter]

this person is confused about something else

[last slate]

oh

what r they confused ab

[merry pewter]

no idea

@vague burrow what are you confused about

[last slate]

finding inverse of coefficent matrix?

haan bhai kya problem hai

[vague burrow]

i did it like

3 times

and each time my answer is different

[merry pewter]

did what

[vague burrow]

findinf A^-1

[vague burrow]

mein bhi indian hu bhai

12th mein hee hu bc rona ara hai

[merry pewter]

ok do you still need help with something?

[last slate]

pata h

mein 11th mein hu 🤣

[last slate]

yeah they can't find the inverse of a 3x3

[vague burrow]

bhai tu mereko paper pe solve krke dede

usse phele vala mera sahi aa gya😭😭

[last slate]

i dont think bro is indian

[vague burrow]

😭😭 can u solve thr whole q and send it cuz im legit half step away from crying

[merry pewter]

no, that's a skill issue, just do it carefully

[last slate]

hahahahhah

[vague burrow]

MEIN

NA

5 BARI APNA PAGE FAR CHUKI HU

😭😭

[last slate]

bhai sun ek kaam kar yeh sab bakchodi chod

[vague burrow]

google pe na hai bhai

[last slate]

youtube video dekh

bsdk common concept hai

[vague burrow]

sab karne ke bad hee yaha ayi hu lmao

[last slate]

3x3 la inverse 💀

puri dunya mein iss par videos h

[vague burrow]

im weak as shit

[last slate]

kya baat kr rhe ho

[vague burrow]

math mein

[last slate]

what percentage are u getting

20%

oh ladki ho 💀

[vague burrow]

hn bhai

ladki hu

math ke ilava sab ata hai mujhe

[last slate]

disrespectfully baat kr liya meine 😭

[vague burrow]

jaldi realise ho gya

[last slate]

sun

[vague burrow]

😭 lmfao

[last slate]

https://metric.ma.ic.ac.uk/metric_public/matrices/inverses/inverses2.html

simple h

[vague burrow]

sad

kha

[last slate]

kya nhi samaj mein aa rha hai

[last slate]

uk

[vague burrow]

who tf made this shit

[last slate]

imperial

college

uk

agar se abhi bhi nahi samaj mein aa rha hai toh i can explain on vc or something

[vague burrow]

try krti hu

galat aaega toh bta dungi

[last slate]

theek hai

tag krdena

padh rha hoon

toh dekhunga nahi

[vague burrow]

okie

[midnight plank]

@vague burrow Has your question been resolved?

[last slate]

@vague burrow have u figured this out

kya chal rha hai bhai

[midnight plank]

@vague burrow Has your question been resolved?

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[waxen silo]

what in the world is the part I highlighted in red? Is it asking for the 2nd derivative or 3rd

[zealous schooner]

It’s asking for the first derivative wrt x and the second derivative wrt y

The order in which you do the three derivatives doesn’t matter

[waxen silo]

so I derive them seperately and combine them together?

[zealous schooner]

You differentiate it once wrt x

And then you differentiate the result twice wrt y

[waxen silo]

oh i see

let me try and lets see what happens

ah got it

thanks kheerii C:

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[misty marlin]

Hi, i just need to be checked. Is this correct? Thank you for giving me your time ❤️

[misty marlin]

the text in book is find sin x/2, cos x/2, tg x/2, ctg x/2

[zealous schooner]

I can't read anything here

[misty marlin]

just second

Is better now

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[midnight plank]

@misty marlin Has your question been resolved?

[misty marlin]

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[zenith finch]

I have to find a polynomial of degree 4 with F(x, y) = 0 and x = r cos(φ) and y = r sin(φ)

[zenith finch]

does this work?

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[twilit field]

so $1<x_1,x_2....x_n>$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is $<x_1, x_2.....x_n>$

?

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is that it?

[civic scroll]

What is 1 in this context?

And F? Is it a field?

[twilit field]

a scalar multiple, I guess

and yes, F is a field

[civic scroll]

It might be a different thing

Sometimes it denotes the identity linear mapping, for example

If 1 is the unit of F, then 1(x_1,...,x_n) = (1x_1,...,1x_n) = (x_1,...,x_n)

[twilit field]

no idea what that means

but is my proof right?

[civic scroll]

I think it's not

You should use the definition of $\lambda x$ when $\lambda\in F$ and $x\in F^n$

[grand pond]

d

[twilit field]

I see

Let me think about this a bit

.close

[midnight plank]

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[gray pumice]

can someone sanity-check my calculation here? I'm not too sure I did this right (matrix multiplication)

[distant heron]

It's good

[gray pumice]

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[cloud vigil]

$a \in \mathbb{R}\
a^2 = a + 1\
log_a(2a + 1) = ?$

[grand pond]

Oğuzhan

[cloud vigil]

Can't use calculators

This is a question from Turkish University Entrance Exam which I've entered today

And this was the only question I couldn't answer

(This is a simplified version of the question, if requested I could send the whole question, but it's most likely not necessary)

[flat veldt]

Are you supposed to simplify this

In terms of log_a

[cloud vigil]

It needs the answer

Also a > 0

Let me type out the entire question with the options

[sage olive]

2a + 1 = 2a^2 - 1 = (sqrt2 a - 1)(sqrt2 a + 1) is all i'm seeing

you could split that product up with log properties but i don't see any nice simplification

i mean if they just want the answer, why not just solve the quadratic for the positive root and substitute for a

[cloud vigil]

$a,x,y \in \mathbb{R^+}\
\
log_a(x),\text{ }log_a(y),\text{ }log_a(x + y)\
$is a series of numbers sorted from smallest to largest and all of its elements are consecutive integers.\
What is the value of $log_a(2a+1)?$\
A) -2\
B) -1\
C) 2\
D) 3\
E) 4

[grand pond]

Oğuzhan

[sage olive]

so where did the a^2 = a + 1 come from in the question?

[cloud vigil]

ax = y
ay = x + y
=> ay = y/a + y
=> a = 1/a + 1
=> a^2 = a + 1

ax comes from log_a(x) + 1 = log_a(ax)

Similar for ay

[burnt flame]

how do you know ax=y???

[sage olive]

yeah i'm confused now

[cloud vigil]

Because they are consecutive

log_a(x) + 1 = log_a(y)

[sage olive]

oh okay

yeah

[cloud vigil]

Therefore this

[cloud vigil]

But how do you proceed from here

[sage olive]

have you tried just solving the quadratic?

[cloud vigil]

Maybe we have to use the fact that the log_a(x) is an integer but I don't see how

[cloud vigil]

It gives a = (1+sqrt(5))/2

I even approximated the value of a in the exam

But it leads to nowhere

[sage olive]

so then 2a + 1 = 2 + sqrt5

[burnt flame]

oh

[cloud vigil]

Yes

[sage olive]

and it's a matter of checking which of those powers you have to raise (1 + sqrt5)/2 to to get 2 + sqrt5

why not just check the 5 options?

[flat veldt]

y/x = (x+y)/y

[burnt flame]

$\log_{a}(2a+1)=\log_{a}((a+1)^{2}-a^2)$

[last slate]

2a+1 = a²+a
log_a(a²+a)
= 1 + log_a(a+1)
= 3

[grand pond]

🫎 Chmoosey 🫎

[cloud vigil]

There are 40 questions and none of them should require you to check the options

[cloud vigil]

I think this is the solution wait

Where did the first come from

[last slate]

add a

[cloud vigil]

Ohhhhhh

a^2 = a + 1
a^2 + a = 2a + 1

Wow

[last slate]

ye

[sage olive]

why is log_a(a+1) = 2?

oh

nvm

[cloud vigil]

a^2 = a + 1

[flat veldt]

Where did value ofv2a+1 come from

[sage olive]

yeah that's clever

[cloud vigil]

@flat veldt ^

Added a to both sides

[flat veldt]

Cool

Smart

[cloud vigil]

Really smart

I wish I had seen that

Thanks again

.close

[midnight plank]

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[reef birch]

I try to understand Universal Approximation Theorem but before understanding the formal proof, I would to understand this:
https://www.desmos.com/calculator/cfvtjusqmq?lang=de

how can it be that adding n_i together approximate the function f ?
Is there a geometrical interpretation or a better intuition?

[pure nest]

yo what..

dawg

[vital sinew]

Please open your own help channel

[pure nest]

can someoe help me.

[vital sinew]

Since ABE and ACD are similar and CD corresponds to BE, what can we say about p

[pure nest]

thats its 4..?

[vital sinew]

Similar, not congruent

[pure nest]

oh

[vital sinew]

In similar triangles, the ratios of corresponding side lengths are the same

In this case, AB corresponds to AC and BE corresponds to CD

[pure nest]

yeah

[vital sinew]

So what can we say about p in terms of the lengths of those other line segments

[pure nest]

its similar

uh

idrk..

thatswhy i came on here

[vital sinew]

Well, we know that AB/AC=BE/CD right

So how can we use that to find p

[pure nest]

how?

[pure nest]

why divide?

or cross multiply

[vital sinew]

#help-49 message

[pure nest]

oh okay

continue

[vital sinew]

We know the values of each of these, so what equation do we get if we substitute those in

[pure nest]

lemme see

ab is 5

ac is 9 i think

be is 4

[vital sinew]

right

And what is CD

[pure nest]

i dont know what cd is..

its not givin

[vital sinew]

It does say

[pure nest]

4? ig bc they are similar

..

[pure nest]

it doesnt say tho

there isnt a umber

oh is it 9

[vital sinew]

p is a number, we just don't know what it is yet

So CD is p

[pure nest]

ohh

i thought u meant what number it is

[vital sinew]

So if we put all of our values into this equation, what do we get

[pure nest]

ahe

ahem..

leme calculate

it

r

rq

7.2?

should be correct

p=7.2

[vital sinew]

Right

[pure nest]

;lets go

ok we found p

what about q

[vital sinew]

So now, can we do a similar thing with q, given that AD=8+q

[pure nest]

yeah

[vital sinew]

Last time we set up this equation #help-49 message, what will we use this time

[pure nest]

hm

p/be= something.

[vital sinew]

Well, q is part of AD

So we want to use the side corresponding to AD

[pure nest]

cb?

de

[vital sinew]

The corresponding side to AD is a side of ABE

[pure nest]

oh

so what formula do we use

p/be=ac/ae?

[vital sinew]

We want to find the length of AD, remember?

And the side corresponding to AD is AE

And we're going to use this fact #help-49 message

[pure nest]

cd/be..

?

😭

[vital sinew]

Right, but what can we set equal to this

[pure nest]

cd/be=ac/ad..

[vital sinew]

Well, the corresponding side to AD isn't AC

#help-49 message

[pure nest]

oh

so cd/be=ad/ae

[vital sinew]

Exactly

Now what can we substitute in for each of those

[pure nest]

whats the vaklue of ad tho

there 8 and then it says q

so ad is 8q?

or just q

[vital sinew]

The value of AD is AE+ED

[pure nest]

so 8q

[vital sinew]

No, 8+q

[pure nest]

if we add tho ts 8q tho, no?

[vital sinew]

No, 8q means 8 times q

[pure nest]

oh yeah

[vital sinew]

So now, what are the values of each of these

[pure nest]

7.2/4=8+q/8

?

[vital sinew]

(8+q)/8, yes

Now can we solve for q?

[pure nest]

uh

yeah

..

[vital sinew]

What do we get?

[pure nest]

one sec

1.7

i think its wrong.

[vital sinew]

Not quite.

What was your first step?

[pure nest]

multiply 7.2 by 8

which is 57.6

[vital sinew]

And then?

[pure nest]

4 multiply 8+q

[vital sinew]

Right

[pure nest]

i got 32q.

[vital sinew]

It's not 8q, it's 8+q

So we have 4(8+q)

[pure nest]

so 32+q.

[vital sinew]

Not quite

[pure nest]

or 32+4q

[vital sinew]

a(b+c)=ab+ac

[vital sinew]

Right

So now what

[pure nest]

divide

ohh

i see what u mean

so it wouldbe

57.6=32+4q

we move 32 to the other side

57.6-32

which is 25.6

so 25.6=4q

25.6/4

[vital sinew]

Right, which is?

[pure nest]

6.4

so q=6.4

[vital sinew]

Exactly

[pure nest]

finallyyyy

😭

thank you so muchhhh

i have a another question if u can help me..

[calm wind]

If we scale one side of a cube by 3, how much do we scale the whole cube by

[pure nest]

i didnt get it

?

[calm wind]

If I have a cube that's 1x1x1 and another cube that's 3x3x3, how many times large is the bigger cube

[pure nest]

6x6x6

[calm wind]

pardon

[pure nest]

😭

[calm wind]

no I mean how many times large is the volume of a cube of side length 3 than a cube of side length 1

[pure nest]

uhm

12

[calm wind]

how do you get 12

[pure nest]

6x2

[calm wind]

why 6x2

[pure nest]

..

thats what i thought of

[calm wind]

what is the volume of a cube

[pure nest]

l cube

cubed

[calm wind]

right so what is the volume of a cube of side length 2, and what is the volume of a cube of side length 6, for example

[pure nest]

uh

2 cubed is uh

2x2x2

16

and 6

is

216

[calm wind]

what is 216/16

wait

16?

what's 2^3

[pure nest]

uhm

8

[calm wind]

Cool what's 216/8

[pure nest]

27

now 27x4? correct?

which is i think 108

[calm wind]

Yes

[pure nest]

what ab tthis?

.close

[midnight plank]

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[pure nest]

.reopen

[midnight plank]

✅

[pure nest]

its been 15min

<@&286206848099549185>

[midnight plank]

@pure nest Has your question been resolved?

[midnight plank]

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[vapid crystal]

can someone just explain to me how to differntiate cosec cubed

[lament fox]

write cosec as sin^-1 and see if that helps

[tribal temple]

Chain rule-

[slender walrus]

power chain

[midnight plank]

@vapid crystal Has your question been resolved?

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[steel smelt]

did i find the derivative of this function correctly?

[slender walrus]

,rotate

[median burrow]

Looks correct

[steel smelt]

thank you

[median burrow]

Wait one small mistake the end

[slender walrus]

no

[median burrow]

(x-3) - 2x = -x - 3

[slender walrus]

you messed up in the last step

[median burrow]

Btw, you can always check your derivatives from:
https://www.derivative-calculator.net/
I used to do it when I was learning derivatives. It also shows you steps so you can see where you messed up

[steel smelt]

tysm!

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[hearty dock]

answer is : pi(e-2)

instead i find pi*e

[hearty dock]

here is my work

i am going to ask this question in the good topics

.close

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[tranquil turtle]

hey how do I solve this

[olive matrix]

draw a right triangle with that angle

how should you label the edges such that cot of that angle is 9/5?

[olive matrix]

sure great

now what is that angle's cosine?

[tranquil turtle]

$\cos{x} = \frac{9}{\sqrt{106}}$

[grand pond]

rynite

[olive matrix]

great

is there anything else you need?

[tranquil turtle]

no thanks 👍 , seems like I haven't read the question correctly from the beginning.

[olive matrix]

.close

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[young jolt]

Where did I go wrong ?

Oh nvm I figured it out

[dull yoke]

Yo where u been man

Haven’t seen u in helpers lounge in so long

[young jolt]

Oh man , what can I say, personal issues to

Yk

[dull yoke]

Ah

hope it gets better

[young jolt]

Now I’m better 😌

Thanks

[dull yoke]

👍

[young jolt]

.close

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[twilit field]

so to start

why is multiplication defined that way

[subtle blaze]

They just decree it to be so

[twilit field]

i feel it's a typo

that feels sus

[hard shard]

no

[subtle blaze]

Why do you add vectors entrywise normally?

[twilit field]

oh right

[subtle blaze]

You just decree that’s how addition is done

[twilit field]

I know, but using the definition of multiplication for complex number, there's a + , right?

[subtle blaze]

We don’t know what a complex number is

[twilit field]

ah yes, that;s how I approach this

[subtle blaze]

[twilit field]

going be be hard to forget everything I know

[subtle blaze]

That’s unfortunately how you gotta start

[twilit field]

[subtle blaze]

Yep, that’s the whole idea of these problems

To get you to unlearn and start from the building blocks

You get to now build all the intuition from logic

[olive matrix]

pretend it's a game

like a puzzle game

[west iron]

Multiplication isn't necessarily "defined" that way but rather it can have any definition that satisfies those properties

[subtle blaze]

I whole heartedly think it’s better to start with some beginner proofs in group theory before doing vector spaces

Vector space axioms look confusing

Group axioms are much simplier to look at

[olive matrix]

i had no trouble with vector space axia when i was in linear

[west iron]

that makes sense, I never truly understood the abstract parts of linalg until after group theory

[olive matrix]

but i think i easily got the "these are the rules you can play by and no others" thing

[twilit field]

okay, so let me start by proving the identity

it's trivial to show that $0 \in V_s$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard shard]

thats the plural?

[twilit field]

by setting a=b=0

[subtle blaze]

Have you proved the axioms from the start

[twilit field]

they're axioms

how can I prove axioms

[subtle blaze]

I mean show that the first few axioms hold

[twilit field]

I was taught that axioms don't have proofs

[subtle blaze]

For the given definition of addition and multiplication

[west iron]

For a specific vector space

?

[subtle blaze]

I worded that poorly my bad

[subtle blaze]

Have you shown this addition is associative

[twilit field]

I'll start with that

as I've said in the past, I work in help channels

[west iron]

ah prove that V_C satisfied the axioms I see

To prove that something is a vector space you prove that it satisfies the axioms

[twilit field]

ok, so let $u=a+ib$, $v=c+di$ , $w= e+fi$

[subtle blaze]

Yeah let him try it out

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

That + sign is fake

It has no meaning

[twilit field]

so we have $a+ib+ (c+e)+(d+f)i$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

Woah you’re mixing up + signs that matter and don’t matter now

[twilit field]

I don't follow

[subtle blaze]

The + in a + ib has no meaning

[twilit field]

why

[subtle blaze]

“a + ib” is 1 symbol that represents the ordered pair (a, b) ∈ V x V

[twilit field]

I still don't follow

[subtle blaze]

You can alternatively see the definition of addition on $V_C$ as

\begin{align*} + : (V \times V) \times (V \times V) &\to V \times V\ ((a, b),(c, d)) &\mapsto (a\textcolor{red}{+}c, b\textcolor{red}{+}d)\end{align*}

[grand pond]

Frosst

[twilit field]

hmm

okay

ooof this is so hard

[subtle blaze]

The question now is… what is this red plus?

[twilit field]

an operator

[subtle blaze]

Where does it come from

[twilit field]

no idea

[subtle blaze]

Hint: ||look at the very top of your question||

Every word is exactly written as needed in these questions

[twilit field]

from the fact that $V$ is a real vector space?

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

You assume no more, no less than you HAVE to

[subtle blaze]

What does this mean?

[twilit field]

going to be hard to get used to

oh

they have defined (u,v)

[subtle blaze]

What’s special about V

[twilit field]

nothing

[subtle blaze]

It’s more than just a set

[twilit field]

oh

right

so it satisfies the properties of vector spaces

[subtle blaze]

It doesn’t

It’s a set

[twilit field]

I'm really confused

[subtle blaze]

The 3-tuple (V, +, • ) is a vector space

[twilit field]

oh

[subtle blaze]

In particular, it looks like $(V, \textcolor{red}{+}, \textcolor{red}{\cdot})$ is a vector space

[grand pond]

Frosst

[twilit field]

why is the + and $\cdot$ important

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

It’s the operations

If you don’t tell me how to add and scalar multiply on my set V, I don’t have a vector space structure

[twilit field]

I see

[subtle blaze]

V is a set by itself

We can add things to it and give it more structure

[twilit field]

so each set can have different deffinitions opf addition

[subtle blaze]

Of course

[twilit field]

so v+u could be defined to always be 1?

[subtle blaze]

Absolutely

[twilit field]

okay

[subtle blaze]

But then perhaps you may not have a vector space…

[twilit field]

sorry for teh stupid questions

[subtle blaze]

These are not stupid questions at all

[twilit field]

{0,1} is a vector space, no?

[subtle blaze]

No that’s a set

You need to tell me how to add and where I can find scalars and how I can multiply them as well

[twilit field]

I see

[twilit field]

back to this

$a+bi +(c+di+e+fi)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

No can do

[twilit field]

not sure

[subtle blaze]

You’re still mixing up the fake + with the real +

[subtle blaze]

Maybe it’s a bit hard to think about it so I’ll say it

The red plus is inherited from the vector space structure of V

By construction now, the new + on V_C inherits all the properties of the red +

[twilit field]

hmm, okay

[subtle blaze]

It happens because you’ve defined the + on V_C by the red + on V

And we know the red + is associative and commutative

[twilit field]

hmm,okay

[subtle blaze]

Why do we know this?

[twilit field]

from teh axioms of vector spaces

[subtle blaze]

Yeah, from V being a vector space

Well, they should’ve been more explicit

[twilit field]

so this give me $(a+c+e)+i(b+d+f)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

Let $(V, \textcolor{red}{+}, \textcolor{red}{\cdot})$ be a real vector space

[grand pond]

Frosst

[subtle blaze]

Yes

[twilit field]

and this proves associativity, right?

[subtle blaze]

I think you should work with ordered pairs

Rather than the a + ib form

Because the + in a + ib is fake

You may get confused which is which

[twilit field]

okay

now to prove commutativity

[olive matrix]

you need my \red \yellow \purple \green \blue macros

[subtle blaze]

Tbh you didn’t really do associativity

[subtle blaze]

Probably do 😭

[twilit field]

oh right

$(a+bi+c+di)+e+fi= (a+b+e)+(b+d+f)i$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

let $u, v, w \in V_C$, and $u = (a, b),, v = (c, d),, w = (e, f)$

[grand pond]

Frosst

[subtle blaze]

\begin{align*}u + (v + w) &= (a, b) + ((c, d) + (e, f)) \&= (a, b) + (c + e, d + f) \&= (a + (c + e), b + (e + f)) \&= \textcolor{blue}{((a + c) + e, (b + e) + f)}\&= ((a+c), (b+e)) + (c, f) \&= ((a, c) + (b, e)) + (c, f)\&= (u + v) + w\end{align*}

a proof of associativity looks like this

[grand pond]

Frosst

[twilit field]

hmm, I see, thanks

[subtle blaze]

this is the proof written is excruciating detail

you should get used to doing them in excruciating detail at the beginning, then once you are comfortable with it, you can start ommiting things not because they aren't important, but because you will be able to do them if you wanted to

the blue line is where i used the associativity of the red plus

[twilit field]

so a. proof of commutativity would looks something like this when written in detail ?

[subtle blaze]

yes

it should

[twilit field]

$v+u=(a,b)+(c,d)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$v+u=(a+c,b+d)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

u+v=(c,d)+(a,b)

[subtle blaze]

what

no

[twilit field]

$u+v=(a+c,d+b)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

woah what

how did the v and the u swap places on the left

[twilit field]

I'm proving v+u=u+v

[subtle blaze]

you dont really want to do this

you want to start with u + v

then use the definitnion of + to get to v + u

[subtle blaze]

if you do it this way, this line is incorrect

it's not wrong but we dont know this equality is true

(it is in fact true but you need to prove so)

[twilit field]

$u+v=(c+a, d+b)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

that is indeed true

[twilit field]

so now what?

[subtle blaze]

good question

you have $u+v=(c+a, d+b)$ and $v+u=(a+c,b+d)$

[grand pond]

Frosst

[subtle blaze]

what did you want to prove

[twilit field]

v+u=u+v

[subtle blaze]

ok what do we need to show then

it sounds really obvious but you need to say it

[twilit field]

they are equal

as addition is commutative

[subtle blaze]

if in doubt, look at this and think to yourself how i moved from 1 line to the next

[subtle blaze]

which addition, and how do you know it's commutative

$v\textcolor{blue}{+}u=(a\textcolor{red}{+}c,b\textcolor{red}{+}d)$

these are not the same + signs

[grand pond]

Frosst

[twilit field]

oh, right

the red one is commutative

right

[subtle blaze]

this is gonna be a very common occurence

literally every reasoning from 1 line to another will be "from definition"

because that is all we've got

[twilit field]

going to take a while to get used to this

[subtle blaze]

it will, and dont beat yourself up for it

this is like actual proper linear algebra in the abstract sense

most highschool/first years encounter this in the concrete example of R^n they dont even know there's different + signs

[twilit field]

now to prove the identity element

[subtle blaze]

sure

[twilit field]

$a+bi+0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

=$a+0+bi$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is simpley a+bi

[subtle blaze]

what is 0

where is it from

[twilit field]

okay

0=0+0

I have to use this

don't I

[subtle blaze]

uh

i dont see how that would help you

[twilit field]

$(a+a-a)+bi$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

that does not parse

a + ib is ONE THING

[twilit field]

$a+ib +(a-a)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

if you want to do things to each entry you need to use the form (a, b)

[subtle blaze]

this also does not parse

$a+ib \textcolor{blue}{+}(a\textcolor{red}{+}-a)$

[grand pond]

Frosst

[subtle blaze]

this is what you wrote

but

the blue + is defined between 2 elements of V_C

$a \textcolor{red}{+} - a$ is not an element of $V_C$

[grand pond]

Frosst

[subtle blaze]

i think you should stop using the a + ib form

use the ordered pair (a, b)

[twilit field]

okay

so I have a,b)+(0,0)?

[twilit field]

is this right?

[subtle blaze]

yep

[twilit field]

so that's just (a,b)

I don't really understand the difference though

[subtle blaze]

#「helpers-lounge」 message

ok we're here now

big

[subtle blaze]

show that (0, 0) behaves like the 0 vector in V_C

ie, show that it has the property you want from it

for any v in V_C, v + (0, 0) = v

show this

[twilit field]

it does (a,b)+(0,0)=(a,b)

[subtle blaze]

how do you know that's true

be more explicit

[twilit field]

it;s a 0 vector

[subtle blaze]

we want to prove that

what's $(a, b) \textcolor{blue}{+} (0, 0)$

[grand pond]

Frosst

[subtle blaze]

use the definition of the blue plus here

[twilit field]

using the defn I'm getting (a,b)

[subtle blaze]

no

what does the definition say

you're jumping ahead

[twilit field]

I'm not too sure tbh

[subtle blaze]

it's (a + 0, b + 0) right?

in particular

[twilit field]

yes

[subtle blaze]

$(a\textcolor{red}{+}0, b\textcolor{red}{+}0)$

[grand pond]

Frosst

[subtle blaze]

this is an important step

[twilit field]

okie

[subtle blaze]

we've defined the blue plus by the red pluses

in this manner

now we say, aha, let me pick the 0 vector from V, and put them in the first and second slot of a thing in V_C

im going to temporarily name the zero vector in $V$ by $0_V$ and the zero vector in $V_C$ by $0_{V_C}$

[grand pond]

Frosst