help-13

[sharp pike]

they might not be

[restive oriole]

I lowkey though this matrix wasn't square for a second before I realized I had no idea what's going on

[sharp pike]

because if you look at the simple case where n =2 they're independent shit lol

[dreamy gale]

I was thinking about Gaussian elimation

[mortal grove]

I would say try to row reduce

[dreamy gale]

Yes but how it’s too complicated to me

[sharp pike]

I would try simple cases like n = 2, 3, 4

and try to find a pattern

which will motivate a general solution

[mortal grove]

I agree with that ^

[dreamy gale]

I searched up on internet and find out this is a circulant matrix so vandermonde matrix may help to calculate but the problem is this method haven’t even been mentioned in lesson at all(most importantly it’s just a beginning course for linear algebra) so I think Gaussian elimination to turn it into REF should be the method But when I wanna use it Things got really complicated I don’t even know how to start

[mortal grove]

like Ari said, try messing around with the matrix if n = 4 for example

then see if you can get it in a simpler form using gaussian elimination

[dreamy gale]

Let me try it

[restive oriole]

wait, is this matrix even square?

[dreamy gale]

Yes it’s square

U can see it’s an n x n matrix

[sharp pike]

yes

each row and column is just 1-n

[restive oriole]

aren't there n columns or am I absolutely tripping

[sharp pike]

yes

[dreamy gale]

It’s n x n

[sharp pike]

and there are n rows

[restive oriole]

lowkey this is way outside my brainpower, sorry for wasting ya'lls time.

[dreamy gale]

wait is it 1

1**n

huhhh seems not

[cedar kiln]

@dreamy gale Has your question been resolved?

[cedar kiln]

@dreamy gale Has your question been resolved?

[cedar kiln]

@dreamy gale Has your question been resolved?

[dreamy gale]

.close

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[wet arch]

Need a bit of help with 14) i) also is someone could actually explain the reverse chain rule that would be great

[cedar kiln]

@wet arch Has your question been resolved?

[upper abyss]

Let's say you took the derivative of (5x^2 + 3)^3. What would you get? Why should you care here?

[wet arch]

Uh

Wdym

Oh I’m doing part i

<@&286206848099549185>

<@&286206848099549185> 🙏🙏🙏🙏

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@wet arch Has your question been resolved?

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[tulip stream]

complete the square on denominator

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[spare widget]

hi

[spare widget]

can anybody help

[outer nova]

What geometry do you know?

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[spare widget]

I know all the circle theorums

Im just bad at applying them

[outer nova]

Including cyclic quadrilateral, I presume?

[spare widget]

yes

I see the cyclic quadrilateral there

[outer nova]

What do the opposite angles of a cyclic quad add up to?

[spare widget]

180

[outer nova]

Alright

Now

What can you deduce about the four angles formed at C?

Not related to the cyclic quad

[spare widget]

they add up to 360

[outer nova]

Anything else?

[spare widget]

uhhh

the opposite one are equal to each otger

[outer nova]

yes!

particularly for us we will need to focus on angle DCF and angle BCE

Let's label them both y.

[spare widget]

ok

[outer nova]

Are you aware of the exterior angle theorem for triangles?

[spare widget]

uh no

sorry

I didnt know it existed

[outer nova]

uhh ok

it's not particularly important we can do without it

Observe DCF

[spare widget]

ok

[outer nova]

What can we say about angle FDC from what we've done so far?

[spare widget]

90 degrees?

not syre

sure

[outer nova]

Consider the angle sum property of the triangle

We have DFC=39 and DCF = y

We know the angles of the triangle must add up to?

[spare widget]

180

[outer nova]

So FDC must be?

[spare widget]

wait we dont have FCD yet

how could we know if there to uknown angles

[outer nova]

No, so just express it in terms of y

[spare widget]

oh ok

39+x+y?

[outer nova]

Eh

[spare widget]

im sorry

[outer nova]

FDC = 180 - 39 - y

Can you tell me why?

[spare widget]

yes

[outer nova]

no worries

[spare widget]

141

[outer nova]

Now look at ADC

ADC + FDC = ?

the angles

[spare widget]

180

[outer nova]

So angle ADC = 180 - (141-y) = ?

39 + y, right?

[spare widget]

yes

[outer nova]

In a similar way can you find angle ABC?

[spare widget]

ABC= 180-(133-y)

[outer nova]

i.e., ABC = 47 + y

Notice in both cases the exterior angle is the sum of the two opposite angles in the triangle, this is called exterior angle theorem

but not super important, js a fyi

[spare widget]

ok

[outer nova]

Now you can apply the cyclic quadrilateral

[spare widget]

so we found two angles in the cyclic quadrialteral

did we

[outer nova]

yes

ABC and ADC are opposite angles of the cyclic quad

From here, find y and give me its numerical value

[spare widget]

(47+y)+(39+y)=180

47+39 is

70+16

86

180-86

94

94/2

47

y is 47

is it?

oops

I did 94/2

wait my brain glitch

nvm

now what

uh

[ancient lodge]

Observe that DCF and x add to 180

[spare widget]

Ohhh

ok

x is 133?

yesss

ty for the help

I was struggling on that questions when practicing for test

I never knew how to solve

now I know

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[spare widget]

how do I clowe

oh pk

.close

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[wary idol]

ammonia or hydrogen sylphyte has rotten egg smell?

[honest field]

hydrogen sulfide but this is a maths server lol

[wary idol]

ty

i cant find a good science server

[honest field]

look at #old-network

there's one for chemistry

[wary idol]

they aren't active like this server is

.close

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[nimble kernel]

Did I do this correctly

[trim valley]

yep

[nimble kernel]

Chill

I know I'm good at math but during the last school year my ADHD was kinda stunting me so I'm just filling in the gaps of what I'm unsure of

[trim valley]

you can close the chat now

[nimble kernel]

.close

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[lament sedge]

Not following a math step in veratasiums new video

[lament sedge]

he defines the h(N) function as an integral of sum

and then defines s(a,N) as another

but I don't see how s(a,n)^3 = sum of e^(i * 2pi (p1 + p2 + p3))

[fringe scroll]

exponent rules

$e^{i \cdot 2\pi (p_1 + p_2 + p_3)} = e^{i \cdot 2 \pi p_1} \cdot e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_3}$

[lament sedge]

yes

[fringe scroll]

then since its an unordered triplet p1,p2,p3, when summing you can do this

$\sum_{p_1 \in P} \sum_{p_2 \in P} \sum_{p_3 \in P} e^{i \cdot 2 \pi p_1} \cdot e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_3}$

[lament sedge]

yes

and you get the same bionomial coeficants

as when you cube?

[fringe scroll]

$\sum_{p_1 \in P} \sum_{p_2 \in P} \bigg( e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_3}\bigg) \sum_{p_3 \in P} e^{i \cdot 2 \pi p_1}$

[fringe scroll]

no binomial coefficients needed

[fringe scroll]

i factored out the p_2 and p_3 stuff

notice that the rightmost sum is just S(a,n)

and you can keep repeating that

if you know what fubini's theorem is, its kinda like that

[lament sedge]

can you show me one more repeat

with the S(a,n)

[lament sedge]

will search that up

[fringe scroll]

$$\sum_{p_1 \in P} \sum_{p_2 \in P} \bigg( e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_3}\bigg) {\c r\sum_{p_3 \in P} e^{i \cdot 2 \pi p_1}} = \sum_{p_1 \in P} \sum_{p_2 \in P} \bigg( e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_3} {\c r S(a,n)}\bigg)$$

[lament sedge]

oh you can factor out the S(a,n)

and the p_3 exponential

and so you will get another S(a,n)

[fringe scroll]

$$= {\c r S(a,n)} \sum_{p_1 \in P} \sum_{p_2 \in P} \bigg( e^{i \cdot 2 \pi p_2} \cdot e^{i \cdot 2 \pi p_1} \bigg) = {\c r S(a,n)} \sum_{p_1 \in P} \bigg(e^{i \cdot 2 \pi p_1} \sum_{p_2 \in P} e^{i \cdot 2 \pi p_2}\bigg)$$

yeah pretty much

[lament sedge]

over here

[fringe scroll]

i mixed up some indices but you get the idea

[lament sedge]

how did you remove the p3 exponetioal

[lament sedge]

oh ok

yeah

I got it thank you

[fringe scroll]

np

[lament sedge]

.close

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[rustic gyro]

Guys how do I turn this into a squared binomial

[dire geode]

,tex .cts

[wraith dagger]

riemann

[rustic gyro]

So my answer is just (x- 5/2)²?

Why is there a c - (b/2)² I'm confused what's that part

[dire geode]

c is 25/4

The "adding 0" part is important

When you add something to an expression, you also have to subtract the something to keep the expression the same

[rustic gyro]

Ahh okay

So 25/4 - (5/2)² is practically just 0?

okay okay

[dire geode]

Yea in this case it's a perfect square

[dire geode]

But the method is more general when you don't have perfect squares

[rustic gyro]

When do I use this method again?

[dire geode]

No idea I'm not your teacher

[cedar kiln]

@rustic gyro Has your question been resolved?

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[hushed trail]

hey, i dont understand how to factorise this quadratic equation, could you help me?#

[hushed trail]

(x^2-2)^2 - 16

[livid hound]

do you know how to factorise a difference of two squares?

[spring sonnet]

change of variable to x^2

[hushed trail]

im not sure what that means, sorry im new to this

[spring sonnet]

Develop the binomial and you will realize that with the change of variable to x^2 it can be solved as a normal square.

[livid hound]

do you know the factorisation for
$$a^2 - b^2 = \what$$

[wraith dagger]

ℝαμOmeganato5

[spring sonnet]

That's faster xd

[hushed trail]

yeah (a-b) (a+b)

[livid hound]

try applying that here

[spring sonnet]

write 16 as a 4 squared

[normal cipher]

Can you make your original term into a difference of two squares?

[hushed trail]

im sorry, im not understanding this terminology. could you show me what you mean

[livid hound]

$16 = \what^2$

[wraith dagger]

ℝαμOmeganato5

[hushed trail]

yes 4^2

[livid hound]

can you see how what you have
$$\red{(x^2-2)^2} - \blue{4^2}$$
is similar to
$$\red{a^2} - \blue{b^2}$$

[hushed trail]

16^2? or 4^2

[livid hound]

sry, typo

[wraith dagger]

ℝαμOmeganato5

[hushed trail]

so (x^2-2)(x^2+2) - 4^2. is that right?

[lone slate]

Oh god no

Keep it as the difference of 2 squares

And your factorization for (x^2-2) ^2 is wrong either way

[livid hound]

can you see that you have
(something)^2 - (something else)^2

or something in the form
a^2 - b^2

[hushed trail]

ok, i have to go. I'll tackle this another time. cheers for the help anyways

.close

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[torn marsh]

I got both of these questions wrong

I was asked to differentiate

[dusty wren]

What have you done to differentiate it?

If you have f*g you can't just say it's differentiation is f'*g'

,tex.diff rules

[wraith dagger]

Executor (ask on server b4 DM)

[dusty wren]

You can use the product rule as shown above

(for the first one)

[torn marsh]

Isn’t this a sum rule tho?

[dusty wren]

They're in multiplication

x^2 * x^1/2

[torn marsh]

Oh nvm yes

[woven crescent]

why is that even called the sum rule, its just distributing the d/dx lol

no way thats a standardized thing

[dusty wren]

As for the second one

How did you write 4sqrt x as sqrt 4

[nova snow]

i think its more commonly referred to as linearity of the derivative

[torn marsh]

What is differentiation f’*g’

[dusty wren]

?

I'm saying that's the wrong differentiation

[torn marsh]

Did I write that?

[dusty wren]

Yeah

[torn marsh]

Remix it real quick

Im not sure how to apply the product rule could u provide me an example @dusty wren

[dusty wren]

sinxcosx

when you differentiate

cosx * cosx - sinx * sinx

[torn marsh]

I did jot learn that

not

[dusty wren]

Then just combine the two powers of x

x^a * x^b = x^(a+b)

[torn marsh]

this is what we were given

rule number 3

can this be also used?

[dusty wren]

Nah

That can be used in the second one

It's called the sum rule

Product rule is different

[torn marsh]

So this can’t be used in question 7

[nova snow]

why is bro purple

[dusty wren]

Nah

[dusty wren]

:)

[torn marsh]

Cool color

[tropic oxide]

i can see like 2 different mixups going on in q8

and for q7 you don't really even need product rule unless you like algebraic suffering

[dusty wren]

Yeah I told to combine powers

[tropic oxide]

uh so where are we at

[torn marsh]

I’m just confused

With a 7

[tropic oxide]

ok do you want to go over one or both of these questions from the beginning then

[torn marsh]

Q7

[tropic oxide]

ok Q7

[torn marsh]

yeah sure

[tropic oxide]

$f(x) = x^2 \sqrt{x}$

[wraith dagger]

Ann

[tropic oxide]

by far the least painful way of doing this is by doing some algebraic magic to the function BEFORE you take its derivative

and by algebraic magic i mean rewriting $x^2 \sqrt{x}$ as $x^{5/2}$

[wraith dagger]

Ann

[tropic oxide]

do you want this explained in more detail Y/N

if you think you might need it explained later then say yes now

i dont wanna go back to it otherwise

[awaiting response]

[torn marsh]

Oh so the exponents get added

I forgot about that

don’t explain it now Ill try the q again

[tropic oxide]

ok

[torn marsh]

Give me 3 mins

[tropic oxide]

one thing i need to mention though

don't equate a function to its derivative

the line where you actually do the differentiation should be clearly distinguished from you messing with f(x) itself

[torn marsh]

Oh yeah true

[tropic oxide]

writing things like $f(x) = x^2 = 2x$ is just no bueno at all; you should write it like:

$f(x) = x^2 \ f'(x) = 2x$

[wraith dagger]

Ann

[torn marsh]

Should’ve been f’(x)

[tropic oxide]

but ok reattempt Q7 and send your work here

[torn marsh]

Is this correct?

[tropic oxide]

notation has an issue still

[torn marsh]

what’s the issue?

[tropic oxide]

only in the very last line do you write what the derivative is

like your answer of $\frac{5}{2} x^{3/2}$ is correct but you fumbled the notation and work layout

[wraith dagger]

Ann

[torn marsh]

How is meant to be

[tropic oxide]

let me get a paper and write it out

[torn marsh]

Oh so from the third line, we equate it to f’(x)?

Since we actually worked it out

[tropic oxide]

the 4th.

[torn marsh]

So in the final answer?

[tropic oxide]

well in this case yes

but do NOT infer "the symbols f'(x) must appear in the final answer only"

[torn marsh]

Wdym

so why have they not done it in the last line with these examples

[tropic oxide]

i think you either misunderstood me or read only 50% of what i wrote

i specifically told you NOT to imagine that there's some rule which says the symbol f'(x) must appear only at the end

f'(x) will appear wherever is contextually appropriate

[torn marsh]

Oh ok so why is it

Appropriate in my case

[tropic oxide]

you will not like the literal answer to that question

so i will instead say dont overthink it

@torn marsh you still here?

[torn marsh]

yes

[tropic oxide]

ok do you have anything else to ask about Q7 Y/N

[torn marsh]

Y but specifically about the f’(x)

[dry bluff]

Your issue is you're assuming f(x) and f'(x) are the same. They are not.
You're not understanding the derivative as an operator, something that takes in an input (a function f(x)), and transforms it into a completely, different output (the derivative, or f'(x)).

Ann has been trying to explain to you that f(x) = x²√(x) = x^(5/2) will never be equal to f'(x) as they are fundamentally different.

I'll give an example. Say f(x) is a food and f'(x) is when you put the food in the blender.
f(x) = banana
f'(x) = banana puree

Your error is just like saying the banana is the same as the banana puree, when it's obviously not.

[tropic oxide]

ive kinda been saying the same thing

[tropic oxide]

ok then re-ask your question

[dry bluff]

GOD that was an embarassing analogy

[dusty wren]

banana puree

[tropic oxide]

@torn marsh

[torn marsh]

yes I was reading the above

[torn marsh]

so how am I meant to write it

[dry bluff]

well, do you see the error in your working?

[tropic oxide]

i showed you how i'd write the solution to Q7 following your format

[torn marsh]

No I don’t

[dry bluff]

you equated two different things

[tropic oxide]

m8 i pointed it out to you

[dry bluff]

x^(5/2) does not equal 5/2 x^(3/2)

[tropic oxide]

do you want me to bring back the images where i did it

your work contained this error: you wrote f'(x) = [something that ISN'T f'(x)]

does that make sense to you Y/N

[torn marsh]

I get what ur tryna say but I don’t seem to get why it’s an error

[tropic oxide]

writing an equals sign between two things that arent equal

is it not obvious why thats wrong

[nova snow]

its like saying 3=5

[torn marsh]

Oh ok I get it now

But the whole purpose is me tryna find the derivative right

[tropic oxide]

yes sure

[dry bluff]

Do you know what a derivative actually means?

[torn marsh]

So if I equate it to the end that’s just the same issue but at the end

[torn marsh]

yes the gradient of the curve

[tropic oxide]

we're trying to sort out a detail in the process and how to write down your work

not really the time or place to wonder about meaning

[dry bluff]

does it make sense that the gradient function of a curve can never be equivalent to that function itself

[tropic oxide]

e^x would like to know your location

[dry bluff]

LOL okay

[honest field]

e^x:

[dry bluff]

but I'm generalizing

[tropic oxide]

anyway it is like trying to tell someone to imagine the texture of an apple when they're trying to spell the word "APPLE" -- it's unhelpful

[dry bluff]

Okay I'll get out of here

[honest field]

I didnt follow this but what's his problem specifically

[torn marsh]

Yea because we need 2 points on the curve but it can’t be at the same point

[tropic oxide]

his problem is writing shit down on the page in a way thats actually correct and doesnt mix things up

[tropic oxide]

completely irrelevant to the current issue

forget about it

[torn marsh]

yea mb

[tropic oxide]

1. do you have anything else left to ask about Q7 Y/N
2. do you have anything else left to ask about your work Y/N
3. do you want to move on to Q8 Y/N
   (answer each question separately)

[honest field]

yea the way he writes literally leads to mistakes

[torn marsh]

About the same issue if that’s okay cuz I don’t wanna get marksed down

[tropic oxide]

ok you didnt tell me which of my 3 questions you were answering so i have to assume it's about my 2nd

[dry bluff]

From what I thought of it, I thought they had a conceptual or intuitive misunderstanding about the nature of f(x) and f'(x) and how they are different, how else would such issues originate?

[tropic oxide]

from my pov it's a writing issue

[torn marsh]

Yeah the function cannot equal f’(x) because f(x) is just a graph and not the gradient

[honest field]

ye idk at first he just doesnt know which rule to use to solve

[tropic oxide]

so let me summarize this for the like 3rd or 4th time:

when writing your work for taking the derivative of a function, it is very often the case that you do some algebraic magic to it before the actual differentiation step(s), or after the actual differentiation step(s), or both.
for this reason it's important to clearly demarcate in your work where the actual differentiation happens.
and this is why you have to write f(x) = foo = bar = baz ... and then f'(x) = other-foo = other-bar = other-baz ...

[honest field]

then the argument comes to writing issues

[tropic oxide]

@torn marsh does this make enough sense to you that we dont have to return to this later

[torn marsh]

i think it’s better for you to write it on paper with examples, instead of me focusing on words

this is a difficult concept for me to grasp

[void spoke]

hell you can club powers if they are of the same base as they were in your question like x^2 . √x
law of exponents says
x^a.x^b=x^(a+b)
Moreover you can also check product rule which says
If they are two function f and g and you have two differentiate f*g with respect to a variable
you get the answer as f'g+fg'

[tropic oxide]

i'll say this as nicely as i possibly can: you're late to the punch and your repetition of the underlying math doesnt contribute anything

[void spoke]

I like tried to explain in three respects

[tropic oxide]

examples of what

[void spoke]

Sorry I was just trying to help actually I could not read full messages

[tropic oxide]

do you want some more complicated differentiation problem where i could illustrate how this algebraic work is done

@torn marsh

[void spoke]

Examples as you gave above like should we show an example of how to such a question ?

[torn marsh]

yes sure

[void spoke]

Ok

Lemme get 1

[torn marsh]

I’m responding to @tropic oxide

[void spoke]

Ok

[tropic oxide]

...

@void spoke i think it would be better if you let me handle this on my own.

[void spoke]

Yea no problem

👍

[torn marsh]

Thanks for your help tho man

[tropic oxide]

do you have a specific one you'd like me to do or should i make one myself

[void spoke]

Nah man say thanks to Ann

She is helping you

👍

[tropic oxide]

im gonna try and produce work with commentary

[torn marsh]

sure make one ur self

cuz I don’t have any

[tropic oxide]

ok as you say, gimme like 5-10 minutes

[torn marsh]

ofcourse

[tropic oxide]

if my handwriting's not clear at any stage please tell me and i can tell you what i wrote.

[torn marsh]

I like ur handwriting dw

I havnt learned the chain rule yet

[tropic oxide]

blech

well... i think my point about laying out your work still basically stands, doesn't it

obviously you don't have to write any of these comments in green or number your lines in red like i did -- i only did that for illustration purposes

but you see now roughly how to lay out your work in the proper way, right?

[torn marsh]

Yes, I get the gist of it now

[tropic oxide]

ok

do you want to go over Q8 then

[torn marsh]

I’ll attempt it again

for now I’ll close the ticket

.closs

.close

[cedar kiln]

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[quasi flint]

Let A and B be two independent witness in a case. The probability that A will speak the truth is x and the probability that B will speak the truth is y. A and B agree in a certain statement. Find the probability that this statement is true?

[quasi flint]

Help me with this

[tropic oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[quasi flint]

1

[tropic oxide]

ok

so A and B are sitting in separate rooms and dont hear each others answers (and in fact they're just called independent witnesses to reinforce this)

do you understand what it means for A and B to agree on a certain statement? Y/N

[quasi flint]

Y

[tropic oxide]

ok explain in your own words

what does it mean for A and B to agree

[quasi flint]

Well it means A AND B WILL give a common statement

[tropic oxide]

ok thats true but i expected a bit more detail

A and B are either both telling the truth, or both lying.

does this make sense to you

[quasi flint]

Yes

[tropic oxide]

ok

can you work out the probability that A and B would agree in the first place?

the question asks for P(true | agreement) so we're working out P(agreement) as a stepping stone for that

[quasi flint]

I cant think of a way to calculate that

[tropic oxide]

A and B are either both telling the truth, or both lying.

[quasi flint]

So 50 50

?

[tropic oxide]

extremely off the mark

P(A truth) = x
P(B truth) = y
this is stated in your question

[quasi flint]

Yes

[tropic oxide]

your answer will involve the letters x and y

P(agreement) = P(A truth & B truth) + P(A lie & B lie)

do you understand this Y/N

[quasi flint]

X/x+y for a?

[tropic oxide]

there are unfortunately at least 3 things wrong with what you just said

so no, try again. read what i said once more.

A and B are either both telling the truth, or both lying.
P(agreement) = P(A truth & B truth) + P(A lie & B lie)

P(A truth) = x
P(B truth) = y
A and B's responses are independent

dont try to guess

[quasi flint]

Okay but isnt it possible one is telling the truth

While other not

[tropic oxide]

no because then they would be disagreeing...

[quasi flint]

I see

Now i get it

[tropic oxide]

you "get it" as in you need no more help?

or you get it as in you can work out what i told you to & continue with me?

[quasi flint]

No i need help

[tropic oxide]

ok then work this out. calculate P(agreement) in terms of x and y.

[quasi flint]

I meant i just understood the p(agreement)

[lilac flame]

Got real quiet

[quasi flint]

Be back after reading some basic stuff xd

😭

Otherwise i cant calculate

[lilac flame]

Do you know what conditional probability is?

[quasi flint]

I do know the defination

Its when even of A depends on event of B

[tropic oxide]

ok

hold on

lets not get off track

i gave haruki something to do

im waiting for him to actually do it

[quasi flint]

Yup

[lilac flame]

Riyal

[tropic oxide]

i can take you through it step by step

[quasi flint]

I am getting 2xy-x-y+1

[lilac flame]

Hmm theres a sign mistake i believe

[quasi flint]

Typo

[tropic oxide]

xy + (1-x)(1-y) would've been fine too but yes that is correct now

ok

so do you see how to proceed from here?

[quasi flint]

We have to check weather its true but idk how

[tropic oxide]

no we don't have to "check" anything.

we have to find the probability that A and B are both telling the truth given that they are in agreement.

[quasi flint]

So conditional

[tropic oxide]

you should already know that P(both truth) = xy

[tropic oxide]

yes exactly

[quasi flint]

Yes

[tropic oxide]

you should also understand that P(both truth & agreement) = P(both truth) because obviously both ppl telling the truth implies they're in agreement

so you got your numerator and you got your denominator

that's almost it

[quasi flint]

What do u mean by this

[tropic oxide]

i mean it in the most literal sense

you know $P(\text{truth} | \text{agreement}) = \frac{P(\text{truth})}{P(\text{agreement})}$

[wraith dagger]

Ann

[tropic oxide]

and you know P(agreement)

and you know P(truth)

youve already done 95% of the work

[quasi flint]

Okay i got it

So the answer is $xy/(1-x-y+2xy)$

[wraith dagger]

Haruki

[lilac flame]

Yoooo cook

[tropic oxide]

$\frac{xy}{1-x-y+2xy}$

[wraith dagger]

Ann

[tropic oxide]

this is how you make fractions

[quasi flint]

@tropic oxide thanks

[cedar kiln]

@quasi flint Has your question been resolved?

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[upper ruin]

A substitution such as t = e^x really helps

[ember heath]

U van factorize it

[wintry monolith]

Yes

Then u can see a root clearly

Synthetic division

[drowsy obsidian]

Ok

Trying

[wintry monolith]

There is a short cut tho

[drowsy obsidian]

I divided by e^2x on both sides

Then subb e^2x =t

[upper ruin]

No need to do that

[dusty wren]

Why

[drowsy obsidian]

To make it quadratic

Easier to solve

[dusty wren]

Yeah but you're still left with a 1/t^2 on rhs

[wintry monolith]

No just

Put exp as t

Then one root is 1

U can see

[ember heath]

First sub e^x as t and then divide by t^2

[wintry monolith]

Divide the polynomial by

X-1

U will get a cubic

also at the last

Ignore the negative roots

And roots below 1

Cause exp minimum value is 1

[drowsy obsidian]

Yh then i let 1/t + t =y😭

I am trying

[drowsy obsidian]

Idk howw

[dusty wren]

Long division

[wintry monolith]

Learn it

If u want u can also use derivatives

[drowsy obsidian]

Ok

Later

[dusty wren]

t*, but yes

[wintry monolith]

U can also use

Sign change method

But it’s very much a rough idea to get real roots

[drowsy obsidian]

Hmmmm

I got it

By substitution

Thanks guyz!

.close

[cedar kiln]

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[cedar kiln]

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[wind forge]

don't know what to do with the x

[oblique flare]

log(a)+log(b)=log(ab) would be your way to go

[wind forge]

yeah about that

[honest field]

set condition x>3 first

[wind forge]

does the base get added to each other ?

[dusty wren]

Remember to check domain

[oblique flare]

no

[dusty wren]

,tex.log rules

[wraith dagger]

Axe cutter (ask on server b4 DM)

[dusty wren]

First one

[wind forge]

yeah I know

[oblique flare]

the base is always the same

[dusty wren]

The base remains the same

[wind forge]

oh

[oblique flare]

never changing (unless using the base change one)

[wind forge]

wait lemme show u guys

[oblique flare]

Ok

[wind forge]

what to do with the output

[oblique flare]

you got that log_2(x(x-3))=2 yeah ?

[wind forge]

umm

I'm getting log_2(x^2-3x)=2

[oblique flare]

yes going good

[wind forge]

I'm stuck at this

[tropic oxide]

how would you solve log_2(z)=2

[wind forge]

wait

1 I guess ?

[tropic oxide]

no guessing

have you done your reading on logarithms since we last spoke?

[wind forge]

yess

[wind forge]

that's how I made it till here

[tropic oxide]

this is very basic and should never be neglected or forgotten

[wind forge]

wait

which property is this

oh

no

wait

[tropic oxide]

definition of log

[wind forge]

i know this

yeah

this is the basic

I just get confuse with the variables

[tropic oxide]

yeah well that's what practice is for innit

[wind forge]

but z should be 1 right ? cuz 2^1 = 2

Isn't it ?

[tropic oxide]

2^1=2 but z isn't 1

you're confusing the placement

a and b are both 2 while z is z

[wind forge]

how

I don't understand

then what should be z

[tropic oxide]

ok i think maybe i should color code things

@wind forge look at this

stare at it for a minimum of 37 seconds

[wind forge]

37 seconds

wow alr

[tropic oxide]

does it make sense to you now

[wind forge]

yess

very much

[tropic oxide]

do you see how to continue

there will be an important point at the end

[wind forge]

So z is 4

[wind forge]

lemme try on my own first

[tropic oxide]

ok

[wind forge]

although I doubt I can do it correctly

[tropic oxide]

well try anyway and send your work on paper

[oblique flare]

,rccw

[tropic oxide]

so far so good

keep going

[wind forge]

wow

I'm doing it correct ?

[oblique flare]

yes

[tropic oxide]

yes as i said

[wind forge]

so now middle term split ?

[tropic oxide]

it's a quadratic

solve it in any way you want

[wind forge]

k

wait

[tropic oxide]

when you get the roots of the quadratic, STOP and pause

[wind forge]

I'm getting -1/2 and -5/2

[oblique flare]

how did you do it (it's wrong)

[wind forge]

i used the quadratic formula

and inserted the values

[oblique flare]

did you use brackets at least because i think some values got messed up

[wind forge]

well I did it mind

just wrote the formula and final answer

lol

wait lemme do it on paper again

[oblique flare]

do it again and check

[wind forge]

Im getting

+1 and -4 now

[oblique flare]

got one of them right but the other wrong

[wind forge]

which one is right

wait

I'm getting the same again

[oblique flare]

it should be x=1 and x=4 (but there is a restriction we'll cover later)

like you could factor it (the easiest of the three methods)

[wind forge]

but -3-5 = -8

and -8/2 = -4

[oblique flare]

wait i got it wrong (a little brain dead and need sleep)
(x-4)(x+1)=0
x=4,x=-1

[wind forge]

when does the sign change in quadratic formula

[oblique flare]

i think you got it wrong on the -b part
should be -(-3) which is just 3
(3+5)/2=4 and -2/2=-1

[wind forge]

wait

yeah

I got it

-1 and +4

now what

[oblique flare]

last thing tho

one of these isn't a solution

[wind forge]

fantastic

I'm loving this

[oblique flare]

log(z) only works when z>0 z could be any argument so go back to your question and check what x should be greater than

[wind forge]

4 ?

[oblique flare]

x=4 only works because ofc log(x-3) then x should be larger than 3

[tropic oxide]

DON'T DO THAT!

you should NEVER do things in your head

with maybe one exception

[wind forge]

yes exactly my logic

[wind forge]

I'm sorry it's a bad habit of mine

[tropic oxide]

if you are 110% confident that you could be woken up by burglars at 3 in the morning and placed at gunpoint to do something on your first try, without mistakes

THEN you can do it in your head.

otherwise no.

[wind forge]

nice way to put it

[oblique flare]

atp i'll just use a calculator to calculate 9+5 lol

[wind forge]

what should I do with the 4 ?

x = 4

so

just insert the value ?

[tropic oxide]

so these are the correct solutions for the quadratic

now there's one important note

the domain of logarithmic functions

do you know what kind of values x can be in order for log_2(x) to exist?

(if not, you're about to find out)

[wind forge]

I got it guys

its

log_2 4=2

[wind forge]

not 0 ?

x>0

[tropic oxide]

x>0 that's right.

[wind forge]

Oh

wait

oh

[tropic oxide]

the stuff inside a log function has to be positive.

[wind forge]

he said it's wrong

[tropic oxide]

who did

where

[wind forge]

he reacted

[tropic oxide]

yeah cause "not 0" is the wrong condition

the right condition is that x should be positive

[wind forge]

yeah my bad

mmhm

[tropic oxide]

and when you've got multiple logs in an equation they all need to have positive numbers in the argument

[wind forge]

that's why I corrected it

cuz not zero can be -ve

[tropic oxide]

so you need to check x=-1 and separately from that x=4 against the following criteria:

• is x itself > 0?
• is x - 3 > 0?

[wind forge]

alr

[cedar kiln]

@wind forge Has your question been resolved?

[cedar kiln]

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[icy rune]

Does anyone know how to solve this?

[raven anvil]

It kinda scares me

[icy rune]

Me too

[raven anvil]

Is this regex?

[icy rune]

yep

[raven anvil]

I skipped this during my coding tutorial

[icy rune]

i dont think this will be in the exam so im thinking about skipping it too ngl

[raven anvil]

I'm a high school student, skipping this worked for me. Idk about you

[icy rune]

my guess was 1((2+3)* + ((2+3)1)*)

[icy rune]

i think i need to know this one

[raven anvil]

Maybe gemini can help

[clear umbra]

honestly idk if it'll work but try 1(2+3)* + 2(1+3)* +3(1+2)* + $(empty string)

it probably wont tbh

[icy rune]

sadly no, i tried that one before 2

[clear umbra]

yeah i figured

[icy rune]

it says (1(?!1)|2(?!2)|3(?!3))+ not sure what ? and ! are

[raven anvil]

You can ask him/her/pronouns

[icy rune]

ou're asking for a regular expression that describes the language where "11", "22", and "33" are not substrings, without using the special regular expression metacharacters ? or !.

This is a very challenging, if not impossible, constraint for most practical regular expression engines, especially for an arbitrarily long string.

Im skipping this one😰

[south tundra]

If you are comfortable with finding regex of a language accepted by a DFA, it is clearly easier to construct a DFA accepting L and do that

[icy rune]

I constructed a DFA for that i think

But still cant find regex for that

smth like this maybe?

I think this is correct

[south tundra]

Why is every state accepting?

[icy rune]

cuz they dont include 11,22, and 33

that means empty,1,2,3,13,12, etc are also accepting or no

[south tundra]

Every state being accepting means the DFA will accept any string

Oh, what you have is an NFA

[icy rune]

oh yep

mb

[south tundra]

Pretty sure you don’t need this many states though

[icy rune]

yeah could be, this is what came to mind first

but imma be honest, the regex still doesnt seem clear to me

😢

[south tundra]

I can’t really write anything down at the moment, I will be back in 20 mins unless someone else takes over

[icy rune]

okay thx

[quasi merlin]

(1|2|3)((1|2|3)(1|2|3)+)?

Could this work?

[icy rune]

nope dont think os

[quasi merlin]

Ohhh I misinterpreted the task

That is much harder

[icy rune]

yep

[south tundra]

@icy rune Well, remember that you want your DFA to simply detect a pair of same symbols being next to each other and there are only 3 symbols, so you can get away with using just 5 states: one start state, three states representing which symbol was read last and a reject state

[cedar kiln]

@icy rune Has your question been resolved?

[south tundra]

But yeah it's still a pain to convert this

[quasi merlin]

Pretty small DFA tho

[icy rune]

ill just skip this one

ty for ur help though

[icy rune]

aggree

[cedar kiln]

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[cedar kiln]

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[quasi flint]

I have no clue how to prove this

[glossy inlet]

what's alpha

[hidden mural]

isn't Poisson rv discrete

[quasi flint]

No clue

[bronze prairie]

Hi! Here's a tip to set you on the right direction. If you take P(X<n) = sum of probability from 0 to n, notice that if we differentiate this expression with regards to the parameter mu, something interesting will happen.

It's kind of a magic trick! But it will lead to the expression you have here

[glossy inlet]

is alpha meant to be mu or something

tbf that does make it the poisson pmf

And the CDF' = PDF thing holds

[bronze prairie]

Noticably... ||p'(k) = -p(k) + p(k+1)||! So this can help you get the final desired expression

[cedar kiln]

@quasi flint Has your question been resolved?

[cedar kiln]

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[short hatch]

when trying root test i also get 1 (i stopped at the end).
what should i do?
any other test?
maybe i have a mistake?

[flint plinth]

do you know stirling's approximation?

[short hatch]

yep

i used it in one of the pictures

[flint plinth]

ah i see, i didn't even look at the second picture yet

let me take al ook

i agree with your first picture, the ratio test gives you 1, so indeterminate

[short hatch]

also in the second picture it's indeterminate

[
(2\pi k)^{1/(2k)} = \pi^{1/(2k)} \cdot (2k)^{1/(2k)}
]

[wraith dagger]

Koren

[short hatch]

so it's 1

and the k-th root is also one

[flint plinth]

maybe instead of using the root test, try to do a limit comparison test, also using stirling's approximation?

i haven't tried it yet so i don't know if it will work

[short hatch]

but compare it to what

maybe the necessary condition for convergence doesn't work

cause it doesn't go to 0

[flint plinth]

yea if you were to just naively replace k! with the stirling approximation, the terms of the series simplify to sqrt(2 pi k), which suggests that you should be able to show divergence by finding a lower bound that diverges

[short hatch]

ok got it

tysm

[flint plinth]

in particular these bounds should be useful:

[short hatch]

Works

[flint plinth]

nice

[cedar kiln]

@short hatch Has your question been resolved?

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[livid marten]

MNKl parallelogram. Side NK is twice the length of MN. The bisector of M hitting line NK is point A. The bisector of N hitting line ML is point B. If AB=15, find the perimeter.

[livid marten]

so far I’ve been trying to think

[normal cipher]

Youre done arent you

[livid marten]

?

[normal cipher]

Oh wait

Where are you stuck at

[livid marten]

I’m not sure how to find X

[normal cipher]

Are you stuck at relating 15 to x

Okay

[livid marten]

Ok

[normal cipher]

Okay

Label the center of the NMBA as O

[livid marten]

Ok

[normal cipher]

We want to show that NMBA is a parallelogram, more specifically a rhombus, right?

Then were done

[livid marten]

NMBA is not exactly a parallelogram, when I drew it, it just came to look like one 😭

[normal cipher]

But it will be

[livid marten]

Oh rlly?

[normal cipher]

It is supposed to be a rhombus

[livid marten]

Oh

[normal cipher]

Sometimes your drawing gives you clues

[normal cipher]

Can you see why?

[livid marten]

Because NO=OB and MO=OA?

[normal cipher]

Precisely

Were done

The four triangles are congruent

[livid marten]

So ur saying X=15?

[normal cipher]

Namely NOM, MOB, BOA, AON

[normal cipher]

Indeed

[livid marten]

that means the perimeter is 90. But there aren’t 90 in the answers..

[normal cipher]

wat

[livid marten]

The answers I’m given is
A. 30 B. 21 C. 20 D. 25 E. 46

that’s why I’m literally stuck

[normal cipher]

wtf

[livid marten]

I spent the last hour trying to make sense why it isn’t 90 or maybe 60 or smth

[normal cipher]

pmo mb i cant find an answer then

[livid marten]

Alr

it’s ok lol

should I close it yet tho?

[lone slate]

Wait for another helper

[livid marten]

alr

[lone slate]

@livid marten I think you could do some angle chasing, I don't really know

[normal cipher]

perhaps, maybe wait a bit

i cant find anything wrong with my method

it should definitely be a rhombus

[livid marten]

I thought the same as well

At first

[lone slate]

Why would this be true?

[raven shard]

did you transcribe this question correctly?

[livid marten]

Yes

[raven shard]

there is at least one error in this transcription it seems

[raven shard]

really? "The bisector of M hitting line MN" ? at the very least this surely can't be right

[livid marten]

Oh it’s just a letter mistake but the drawing

[raven shard]

yes i am asking if there are more errors in the text you posted

[lone slate]

Hmm

[lone slate]

Did the question provide any diagram for u?

Other than the one you drew*

[livid marten]

yes that’s the diagram it gave

[lone slate]

Hmm

[livid marten]

I’m unable to send the picture because it’s in completely different language

[normal cipher]

its okay

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[livid marten]

Am I supposed to reply the question to it?

[lone slate]

Idk u could do angle chasing mahbe??

[normal cipher]

You can

[livid marten]

MNKl parallelogram. Side NK is twice the length of MN. The bisector of M hitting line NK is point A. The bisector of N hitting line ML is point B. If AB=15, find the perimeter.

[raven shard]

the "answer" to the question as it is written is 90

[livid marten]

What is this question aurghsgsha. Why doesn’t it say 90 on the answers I’m so done

[cedar kiln]

@livid marten Has your question been resolved?

[livid marten]

actually let’s just end it

.close

[cedar kiln]

Channel closed

Closed by @livid marten

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[signal oak]

hello its been a while since I've done multivar calc

[signal oak]

can someone check my chain rule

specifically for the u_xx part ..

[dire geode]

What even is the original question

[signal oak]

original question was a pde

need to change of coordinates y= x-at to get rid of the constant a to solve the pde

[dire geode]

Yea then your second order partial is correct

[signal oak]

since i was changing coordinate systems i needed to uhhahhh use some chain rule and then i realized i forgot how to do u_xx chain rule so i had to reread my old notes

[signal oak]

thank you for checking

.close

[cedar kiln]

Channel closed

Closed by @signal oak

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fringe moth]

Id need help with some math questions

[fringe moth]

What is the value of the definite integral of the function f(x)=e^−x calculated between x=0 and x=1 ?

I am familiar with integrals, in fact I studied them 1 year ago but forgot most of it

I'm not sure where to start

[mental trail]

FTC

find an antiderivative

[fringe moth]

just for curiosity, I dont remember doing this type of step while I studied integrals, is this specifically related to this type of question or its common while resolving integrals?

[mental trail]

Uh, it's common

[fringe moth]

okay thank you

[hollow trail]

that's what you do as a first step for like 90% of definite integrals

[fringe moth]

Ill get on it

Im a bit rusty with math

Okay I understand what is antiderivative

isn't it (e^-x+1)/1+(-x)?

since ∫x^n = (x^n+1)/1+n

[mental trail]

x^n.

x is the variable

n is a constant

you have e^(-x)

[fringe moth]

yes

thats clear

[mental trail]

so

[fringe moth]

so whats wrong

[mental trail]

x is the variable

x^n dx

leads to x^(n+1)/n+1

[fringe moth]

so Im supposed to do e+1?

[mental trail]

no

[uneven tiger]

Int e^x=e^x

[fringe moth]

so e^e^x = e^x?

since ln is always e