help-23

[stuck basalt]

woops

[split kayak]

Consider, these tables were usually tabulated to about the millions in some cases in a time were most people didnt even know how to do basic operations

[flat frigate]

9k

[scarlet storm]

true

[split kayak]

This has to do with the functional definition of logarithms

Such that they are the inverse (aka, undo) exponentiation for the same base

[scarlet storm]

OHH SHIII

THAT'S COOL

[split kayak]

On that note, there are usually some easy values for log base 10

ex.

1000 = 10^3

so log(1000) = 3

You basically dig up the exponent.

[scarlet storm]

hmmmm

[split kayak]

For the sake of clarity
log (without writing base) is in 99% of the cases log_10 / base 10

ln is base "e", or euler's number.

[scarlet storm]

WAAAA THAT'S WHY I COULD'NT FIND IT

[split kayak]

youll eventually see why this base has its own name

[scarlet storm]

uhhh i dont know about these just now

[split kayak]

and log_a is just logarithm base "a"

[scarlet storm]

okay!

[split kayak]

ln stands for natural logarithm. Has to do with the fact that the eulers number is a really reocurring constant in math

probably far more important than pi

[scarlet storm]

is that what the e is on the calculator ?

got it

[split kayak]

~2.71828

its like pi in the sense they both have infinite decimals

[scarlet storm]

ooooo

irrational

[split kayak]

Well, similar to how "log" undoes 10^n

"ln" undoes e^x

and for the general case

log_a undoes a^x

[scarlet storm]

ion get it

[split kayak]

lets say we work with log_3

so logarithm base 3

[scarlet storm]

yea yeah

[split kayak]

now, 27 is the third power of 3

aka, 3^3

[scarlet storm]

yes ye s

kit_BAHHH

[split kayak]

so log_3 ( 27 ) = log_3 ( 3^3 ) = 3

They "undo" exponentiation, since, having 3^n inside a log_3

just equals n

[scarlet storm]

.......

[split kayak]

quick sec

[scarlet storm]

uhhhh maybe am just too dumb 😭

[split kayak]

,, \log_{\color{blue}3}(27) = \log_{\color{blue}3}({\color{blue}3}^{\color{red}3})

Do you agree with this step?

[split kayak]

We are just rewriting 27 into 3^3 aka its prime factors

[scarlet storm]

yes i get it till here

[split kayak]

,, \log_{\color{blue}3}(27) = \log_{\tiny\boxed{\color{blue}3}}({\boxed{\color{blue}3}}^{\color{red}3}) = {\color{red}3}

[split kayak]

Since the base of the logarithm, and the base of the exponentiation are the same

you re just left with the power 3

[scarlet storm]

yes

[foggy fossil]

What's the question

[split kayak]

This can be justified with the core idea of the logarithm making multiplication into addition

[scarlet storm]

ohhhhhhh

[scarlet storm]

nothing, i just wanted to know what logs are..

[split kayak]

log_3(3*3*3) = log_3(3) + log_3(3) + log_3(3)```

And when the base is the same as the number you take the logarithm of, you get 1

log_3(3*3*3) = log_3(3) + log_3(3) + log_3(3) = 1 + 1 + 1 = 3```

[foggy fossil]

Ohh

[scarlet storm]

hmmm hmmm yes yes

[split kayak]

You get to the same result

[scarlet storm]

i get it

[foggy fossil]

Ya it has somebbasic properties w

[scarlet storm]

how is the result the same ?..

3* 3 * 3 is 9 and uhh this gives 3

[split kayak]

,, \log_3(27) = \log_3(3^3) = 3 \
\log_3(27) = \log_3(3^3) = \log_3(3\cdot3\cdot3) = \log_33+\log_33+\log_33 = 1 + 1 + 1 = 3

[split kayak]

I meant the previous method

[scarlet storm]

OHHHH SHITTT

I GET IT NOW

DAMN MB AM DUMB

[split kayak]

There are 4~~(2)~~ rules about logarithms to be honest

,tex .log rules

[scarlet storm]

yes yes i saw these in my book

[split kayak]

Theres a little bit of a problem tho.

Logarithms cannot operate with negative numbers

[scarlet storm]

huh

[split kayak]

well, at least for now in your course

like,

log(-1) doesnt exist

[scarlet storm]

why so

[split kayak]

If you recall that logarithms are like the "inverse" of exponents

[scarlet storm]

but the power can be negative right ?

[split kayak]

there isnt a number n such that 10^n = -1

[split kayak]

again, recall that finding n is essentially what logarithms do. But since there isnt an "n" which satisfies this equation, the logarithms of -1 doesnt exist

[scarlet storm]

hmmmm so -1 just doesnt exist on the table

[split kayak]

And that is true for all values less or equal than 0

[scarlet storm]

ahhh okay!!

[split kayak]

Youll eventually learn that they actually do exist but not in the real numbers :)

[scarlet storm]

so lets take this question

log x 1/4 = -1

[split kayak]

as in

[scarlet storm]

we need to solve for x

[split kayak]

,, \log_x \frac14 = - 1

[split kayak]

Where the base is what youre asked to find

[scarlet storm]

yes

[split kayak]

Well, in this case you prob want to use the identity that makes logarithms into exponents.

[safe radish]

@scarlet storm Has your question been resolved?

[safe radish]

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[dreamy kestrel]

Heya

[dreamy kestrel]

Is my answer.
-1 / 1 - e^1+ ln(2) - 1/x

equivalent to any of those answers?

I wanna know if my answer is correct to any of these

[raven heart]

You mean $\frac{-1}{1-e^{1+\ln(2)-\frac{1}{x}}}$ ?

[flat frigate]

aPlatypus

[raven heart]

@dreamy kestrel

[dreamy kestrel]

yes

[raven heart]

In this case, yeah it's correct

[dreamy kestrel]

Really?

[raven heart]

You have e^ln(2) = 2

[dreamy kestrel]

I asked chatgpt it said it was wrong (though chatgpt is sometimes an idiot)

[dreamy kestrel]

So its correct, they are equivalent

right?

[raven heart]

Yea

[dreamy kestrel]

Thank you. 8 marks secured

.close

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[hoary furnace]

I dont understand the step of taking out the E and how it turns 5.6 into In can someone explain please

[summer coral]

ln(e) = 1, so taking the ln of both sides gets rid of the e

ln(e) is saying, e to the what power will give you e?

[peak estuary]

ln(e^something)=something

by definition of what ln means

[hoary furnace]

uh

[peak estuary]

ln(x) asks the question "to what power do I have to raise e to get x"

so ln(e^something) is something

[hoary furnace]

so what is ln?

ln is just e?

[stuck basalt]

$$\ln e = 1$$
$$\ln x^y = y\ln x$$
so
$$\ln e^{10r} = 10r \ln e = 10r * 1 = 10r$$

[hoary furnace]

but nonetheless how do you from e to the power of 10r to that

did he divide by e?

[flat frigate]

9k

[hoary furnace]

but what operation happened for 5.6 to turn into ln(5.6)

were both sides divided by e?

[summer coral]

ln is the natural log, or just a log with the base "e"

[peak estuary]

the operation of doing ln on both sides

[stuck basalt]

chat gpt took the natural log on both sides

[peak estuary]

the same way we can go from a=b to a^2=b^2 or sqrt(a)=sqrt(b) we can go from a=b to ln(a)=ln(b)

[hoary furnace]

i dont know what sqrt is

[peak estuary]

we apply the same operation on both sides

square root

[stuck basalt]

square root

[summer coral]

$\sqrt$

[flat frigate]

Krish
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hoary furnace]

oh ok

so ln is just division by e?

[stuck basalt]

no

$\ln = \log_e$

[flat frigate]

9k

[stuck basalt]

it’s just log base e

[hoary furnace]

i dont really understand logarithms either

[summer coral]

,tex log rules

[flat frigate]

Krish

[summer coral]

how does that work LOL

someone got it

[stuck basalt]

,ask log rules

[open wedge]

you gotta have a preamable or smth

[stuck basalt]

bruh

[open wedge]

[stuck basalt]

nooo

[open wedge]

try ,tex log rules

[summer coral]

google 👍

[hoary furnace]

cant read math

[open wedge]

anyways

[hoary furnace]

well thanks either way ill figure it out at one point

today is just not the day

[stuck basalt]

chat gpt took the natural log on both sides and then used the power rule to eliminate the e on the right side.

because ln e = 1

[open wedge]

!noai

[safe radish]

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).

[hoary furnace]

no other choice

ai is king

[stuck basalt]

the image Op started the thread with was a screenshot from chat gpt

[open wedge]

[stuck basalt]

here

[open wedge]

ic

[open wedge]

blud thinks AI is going to take over the world for 30 more years

but really

AI could be wrong

[hoary furnace]

beh

[stuck basalt]

how long have you been using ai to help you with math for

[hoary furnace]

way too long

[delicate shore]

why are ducks evil

[hoary furnace]

but again its not like I have another choice I cant read math and I need someone to check answers otherwise I lose grade

[hoary furnace]

they gang rape females

or something like that I forgot already

[stuck basalt]

is it the notation that confuses you

[summer coral]

certain ai's are better than others at math. geometry or anything that involves shapes or pictures is probably the worst thing to ask ai. regardless if you ask you should also know how to check if it gives you the right answer.

[hoary furnace]

I dont even know I just cant read math and convert it into text in my head and without text I cannot understand it

[hoary furnace]

I check if it gives the right answer by doing it myself first

[summer coral]

WolframAlpha is never wrong, its not an ai it is basically a calculator. i recommend you start using that

[stuck basalt]

when you study what do you do do you just read the responses ai gives you

i wouldn’t get too comfortable with chat gpt btw

[hoary furnace]

I usually go through the videos while doing the quiz

and then check with ai

i also ask it to summarize the modules so I have a list of actual things I have to know

[open wedge]

good use mhm

[hoary furnace]

and I also ask it to explain formulas

and stuff like that

I havent had a teacher I actively learnt math from since the 7th grade

I miss pre-pre-pre-algebra

[stuck basalt]

idk i just think it’s bad practice

like
part of learning a topic is also learning how to verify your work

it’s like asking to solve for x in $x+2 = 3$

x = 1
and then you plug 1 back into the original equation to check if you’re correct

[flat frigate]

9k

[summer coral]

back when math class used to be called "math"

[hoary furnace]

well i know that ofc

[hoary furnace]

yeah 😭

[stuck basalt]

yeah but that goes with every math topic you feel me

[open wedge]

how about a quadratic

like if you got $(x - 1)(x - 2) = 0$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

and some guy tells you that $x = 1$ is the only solution

[flat frigate]

1 divided by 0 equals Infinity

[hoary furnace]

i know quadratics its just for me to understand a topic very well I gotta go through a lot of practice questions so it sticks but I dont have the time for that

[open wedge]

you can't just blindly plug in $x = 1$ and say that's the only solution

[flat frigate]

1 divided by 0 equals Infinity

[hoary furnace]

quadratic formula gives 2 solutions

[open wedge]

sometimes

[hoary furnace]

or 1

or 2 imaginary ones

or something like that

[open wedge]

ya

[hoary furnace]

whats it called the determinant

[open wedge]

yeah

$\Delta = b^2 - 4ac$ decides them all

[hoary furnace]

yeah

[flat frigate]

1 divided by 0 equals Infinity

[hoary furnace]

the holy triangle of death

[stuck basalt]

part of learning quadratics is learning the zero product property

if ab = 0 then either a = 0 or b = 0
so it follows from learning about this that you have to account for both possible cases if a = 0 or b = 0

[open wedge]

yeah

i agree so

[hoary furnace]

Im good at quadratics

just not logs I started learning them yesterday

[broken forum]

[open wedge]

imagine giving $x^2 - 3x + 2 = 0$ and some guy $x = 1$ is the only solution, you trust them because you plugged in while you don't know how to factor

[flat frigate]

1 divided by 0 equals Infinity

[stuck basalt]

but i mean like the reason i say don’t get too comfortable with AI is because as you move on to more advanced topics it just makes trivial mistakes where if you don’t have a certain intuition you may not notice it and it’ll mess up your entire work.

like it makes really, really, really dumb mistakes sometimes

[open wedge]

agreed lol

[hoary furnace]

yeah

[open wedge]

this goes for all subjects btw

[hoary furnace]

but thats why i learn it first I use ai as a study tool

[open wedge]

especially LITERATURE

[hoary furnace]

and as a check

I dont like to use it as answer giver unless im really fucked on time

[broken forum]

uni students are really fckd on time every second day :/

[stuck basalt]

one time i told ai i needed to wash my car. i told it the car wash was 1000m away and asked it if i should walk or drive to the car wash

it told me to walk

[open wedge]

it's not like it's dumb

it just thinks that you're broke

💀

[stuck basalt]

i recommend studying the topics in depth and knowing how to check your work

[hoary furnace]

no time

[stuck basalt]

once you understand a topic
then you understand what to expect when you solve a problem

if you know what to expect when u derive a solution
you will know how to verify your work yourself

[hoary furnace]

yeah but i dont trust myself

and I dont trust ai either

so i just do best I can and then force ai to do best it can

[foggy fossil]

But verification wastes time in competitive exams ig

Ai isn't reliable for maths

[hoary furnace]

I know

ok thanks for help everyone

good luck

.close

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[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[grim plover]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[quiet plume]

I'll close this one for now since OP opened duplicates.

.close

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[plucky marsh]

I've been given this problem in Number Theory : Given a finite set P of prime numbers, prove that there exists a strictly positive integer x that can be written in the form a^p+b^p (with a and b strictly positive integers) for all p∈P but that cannot be written in this form for any prime p outside of P. I was able to prove existence but i'm really in lack of ideas to show uniqueness, do you have hints or do you know some tools / theorems which can help ?

[safe radish]

@plucky marsh Has your question been resolved?

[distant merlin]

Study the structure of (Z/q^2Z)* to find residues that q-th power sums cannot hit then use CRT to create an x that lands on a forbidden residue for ever prime

every*

[safe radish]

@plucky marsh Has your question been resolved?

[safe radish]

@plucky marsh Has your question been resolved?

[safe radish]

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[faint glade]

Am I on the right track here?

[tardy mango]

you misread your z as a 2

[faint glade]

oops

long day

I need to solve the equation for y

Now I get $5x^2z-5xy+3yz$

[flat frigate]

Vortac

[faint glade]

how do I go about isolating y?

I don't want the answer, but how to think about it

[tardy mango]

moving all of the terms with y to one side would be a good idea

[faint glade]

So $5x^2z=5xy+3yz$

[flat frigate]

Vortac

[faint glade]

Then would I divide?

$\frac{5x^2z}{5x+3z}$

[flat frigate]

Vortac

[normal moss]

Why + 3yz

You had $5x^2z - 5xy + 3yz = 0$

[normal moss]

@faint glade

[faint glade]

Don't I need to move y to one side?

[normal moss]

Yes

The terms with y

But you moved incorrectly

$3x + 4 = 0$

[normal moss]

$3x = -4$

[normal moss]

Minus 4, not +4

(We subtract 4 on both sides)

Adding / subtracting on both sides is what moving really is

[faint glade]

Where are you getting the 4 from?

[normal moss]

It is an example

[faint glade]

ohh

[normal moss]

When moving terms, the sign changes

[normal moss]

You didn't change the sign of 3yz

[faint glade]

So first, $-5xy+3yz=-5x^2z$?

[flat frigate]

Vortac

[faint glade]

then, $y(-5x+3z)=-5x^2z$?

[flat frigate]

Vortac

[normal moss]

Sure

[faint glade]

Then I do $\frac{y(-5x+3z)}{(-5x+3z)}=\frac{-5x^2z}{(-5x+3z)}$?

[flat frigate]

Vortac

[normal moss]

Yes

[faint glade]

ahhh okay

thank you!

.close

[safe radish]

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[normal moss]

Also

Probably useful to write restrictions

$x \neq 0, y \neq 0, z \neq 0, 3z - 5x \neq 0 \implies z \neq \frac{5}{3}x$

[normal moss]

But no problem 🙂

[safe radish]

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[lone shuttle]

Hey there I was wondering how to solve this Riemann-Stieltjes integration problem, any help would be really appreciated!

[lone shuttle]

We have already shown these two results before which I think are needed for this

The professors solutions are as follows but I really don't understand them:
Let $k_r \coloneq \lfloor \frac{-1}{r}\rfloor$, then $k_r \le \frac{-1}{r}<k_r+1$ i.e. $-\frac{1}{k_r+1} < r \le -\frac{1}{k_r}$, and $\alpha(x) = k$ if $k\le -\frac{1}{x}<k+1$, i.e. $-\frac{1}{k+1} < r \le -\frac{1}{k}$, $\forall k\in\mathbb{Z}$. It follows that:
$\int_{r}^{1}x^pd\alpha=\int_{r}^{-\frac{1}{k_r}}x^pd\alpha + \int_{-\frac{1}{k_r}}^{-\frac{1}{k_r+1}}x^pd\alpha + ... + \int_{\frac{1}{2}}^{1}x^pd\alpha$
By 2, we get that this is equal to
$\sum_{k=1}^{-k_r}\frac{1}{k^p}$, which converges iff $p>1$.
I really dont understand how use used result #2 here (like which value of $s$ did he use).

[flat frigate]

dan

[lone shuttle]

also how does k_r \le -1/r < k_r + 1 even imply -1/(k_r+1) < r \le -1/k_r shouldn't the inequalities be the other way?

[safe radish]

@lone shuttle Has your question been resolved?

[lone shuttle]

.close

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[humble egret]

why is sine used here instead we should be using cos

[opaque fern]

Both are sinusodial waves

Does it really matter

It's just a 90 degrees phase shift

[humble egret]

i can give u the whole problem can u help?

[opaque fern]

For any possible stsrting point you could just write
[
y = a\sin\8{\4{2\pi}\lambda (vt-x) + \varphi}
]

[opaque fern]

Maybe

[opaque fern]

Not that well hearsed in acoustics but aight

Whats your issue

[humble egret]

ok first thing i dont get that we are using the sine component still another thing that are notations all fucked here im so confused

[opaque fern]

Can you explain why not use sine here

[humble egret]

it should clearly be horizontal oscillation

for sound wave is it not

[opaque fern]

Okay?

[humble egret]

if so displacement should be given by the eqn

x=acos(wt+phi)

[opaque fern]

Like there is no notion of 'vertical' or 'horizontal' when using either of those functions on their own. The equation is just calculating magnitude NOT a particle's motion in space

[humble egret]

calculating magnitude either way is calculating some value of x or y so it is calculating the position no???

[opaque fern]

The x here isnt meant to tell you the particle's current position

[humble egret]

ys

ys

sys

bbut it does say for y

tho

[opaque fern]

Yes but in a relative sense

[humble egret]

hheh

wth

wdyum

😭

[opaque fern]

I am trying to tell you that using either cos or sin does not matter because the difference between them is just a phase shift (cos(theta) = sin(90 - theta))

Phase shifts only result in a horizontal shift, which in time, is equivalent to you recording the function starting from different states (I.e., it achieves its maximum displacement at t = 0, when it could very well be its minimum if you start at a later time)

[humble egret]

now i think abt it

y= sin(x) for horizontal displacement is same as y= cos ( x) by now sin(90-x) nono sorry aghhhh

leave it

maybe ill see my teachers

thanks fo rthe help and taking ur time tho

.close

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[distant storm]

In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.
What is the probability of not getting a prize if you buy (a) one ticket (b) two
tickets (c) 10 tickets.

The method I used was 1/1000 is the probability of getting the prize so thats the ans for a, so for b part you have 2/1000, and for c likewise ur having 10/1000. so just 1-P(x) to get the ans

[distant storm]

but my book uses combinatorics and says for C part it is 9990C10/10000C10 etc etc and the answer is kinda the same of what i got

but its not EXACT, so i asked chatgpt (im sorry) and it said my ans wouldnt rlly be right?

am i missing something?

[glossy dock]

P(no prize) = 9990/10000 = 0.999 so thats right

[clever tinsel]

you're assuming that the tickets you bought are 'with replacement'.

[glossy dock]

saying "2 tickets ⇒ P(prize) = 2/1000" is treating the probability as if it just adds up linearly

[clever tinsel]

remember that after considering your first ticket as a loss, you only have 9999 tickets left to compare against.

and so on.

[glossy dock]

this is just the expected number of prizes you win, not the probability of winning at least one

which is why u got an ans thats kinda there but not really

just think about it, for 2 tickets, you want the probability that neither of your tickets is among the 10 winning ones. Out of C(10000, 2) equally likely pairs of tickets you could have bought, C(9990, 2) of them consist entirely of losers

so im guessing the book ans was 0.998001

or around there

9980009

[distant storm]

yeah

[glossy dock]

yeah cause P(no prize, 2 tickets) = C(9990,2) / C(10000,2) = (9990·9989)/(10000·9999) = .998001

[distant storm]

but why doesn't my method work?

what assumption of linearity am i making?

[glossy dock]

think about it

if you bought 1000 tickets, your rule would give P(prize) = 1000/1000 = 1, guaranteed. But that's obviously wrong since you could buy 1000 tickets and still have all 10 prizes land on the other 9000

so ur logic breaks down at n=2

[clever tinsel]

you're assuming that there are still 10000 tickets to compare against after your first n tickets.

[glossy dock]

or as the other person said ur assuming with replacement

aka replacing the ticket back into the pot\

[distant storm]

oh i see. still kinda confused but I do get what's happening with the thing im doing.

ty yall

ill just stick to combinatorics for everything 💔

.close

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[distant storm]

.reopen

[safe radish]

✅ Original question: #help-23 message

[distant storm]

oh wait, mind helping me with this one too

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend
are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

I think I used the same logic of linearity here perhaps, assumed section A and B, P(A)=0.4 and P(B)=0.6, so a) 0.4x0.4+0.6x0.6, and with b) 0.4x0.6x2!

[glossy dock]

so what is ur issue

[distant storm]

Uh the answer is apparently wrong 😭 but I don't get why. Seems pretty basic to me to add the probabilities

[glossy dock]

the two placements aren't independent cause once you're placed in a section, there's one fewer seat there for your friend

[distant storm]

oh oops

[glossy dock]

P(you in A) = 40/100, and then P(friend in A | you in A) = 39/99 cause there is only 39 seats left in A out of the 99 remaining students

[distant storm]

💔 oh well time to leave maths

tysm

ill redo it

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[outer yoke]

I want plenty pings

[opaque fern]

<@&268886789983436800>

[lilac crane]

what?

this does not seem like a math problem

only use help channels for math problems

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[twin prawn]

for number 4, the clue there is that

its a factorial

1, 2, 6, 24, 120

i genuinely dont know where to begin is the problem

i guess theres a + 1 at the tail end of the formula

its a geometric sequence and its initial term is 1

its prolly n! + 1 but idk if im allowed to use that

[fathom jewel]

yes that seems right

[twin prawn]

oh we're allowed to use ! ?

[fathom jewel]

how else would u describe it

[twin prawn]

idk either hahahah

but thanks, turns out i was overthinking it

[fathom jewel]

you dont wanna use a product

[twin prawn]

for number 1) is it

$a_n = a_{n - 1} + 6*(3^{n - 1})$

[flat frigate]

Maladroit

[fathom jewel]

,w RecurrenceTable[{f[n] == f[n - 1] + 6*3^(n - 1), f[0] == 5}, f[n], {n, 1, 5}]

[twin prawn]

assume a_0 = 5

[flat frigate]

Results provided by WolframAlpha

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

Requested by 4nzon

[twin prawn]

hell yeah

[velvet nebula]

tbf you can make the sequence to be anything you want after the ...

[twin prawn]

wdym

[velvet nebula]

5 11 29 83 1 1 1 1 1 ... is a valid sequence

[twin prawn]

.close

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[vague phoenix]

11111 ☠️

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[fervent pulsar]

how is PI -2 for b

[raven heart]

who says it's -2 ?

@fervent pulsar

[fervent pulsar]

mark scheme

😭

i thought it was -2x

[raven heart]

Ah yeah right I misread the equation

-2 works indeed

Well the easiest thing here is to search for v=constant, cause your RHS is a constant and v shows up in the LHS

And if you search for such a sol, you'll get v=-2

[fervent pulsar]

but how come

[raven heart]

How come what

[fervent pulsar]

like isnt it like idk how to explain

like if u had 2x^3 on the right hand side

ud do

ax^3+bx^2+cx+d

as ur PI

idk what u do if its a constant

well

nvm i see now

but

yh yh

i get it

thank you tho

[raven heart]

<@&268886789983436800> spam

[near sky]

<@&268886789983436800>

another

[fervent pulsar]

.close

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[primal gazelle]

This below is the probability of the complimentary event, but idk how to compute this
$\prod_{i=1}^{\infty} \left( 1 - \prod_{j=1}^{k} \frac{1}{(ki+j)^\alpha} \right)$

[flat frigate]

k

[primal gazelle]

well not exactly the complimentary event, but the probability that in each separate block of k trials we get at least one fail

in separate intervals

[safe radish]

@primal gazelle Has your question been resolved?

[round egret]

I think they just want you to leave it in product form

[primal gazelle]

oh

yeah possibly

[round egret]

yeah computing it looks too difficult

I guess if you want to, you could use the fact that
$\ln(\prod a_i) = \sum \ln(a_i)$ to approximate it

[flat frigate]

MxRgD

[round egret]

so turning the product into a sum

[primal gazelle]

thats what I was thinking about too, but then it still looks kinda bad and I didnt want to use approximations

okay then imma just focus on other probability stuff this is probably not the point of the exercise anyway

thank you

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[stiff nest]

so confused

on

part

aii

f(x) > -16

[simple gazelle]

what have you tried

[stiff nest]

g(x) < 8

drawing it

lemme show u

@simple gazelle

[simple gazelle]

since it's f(g(x)), try finding when g(x) is less than 0

since f(x) isn't defined for anything above 0

this will find you the domain of the function

[stiff nest]

so

x < 24

[simple gazelle]

x > 24

but you got it

[stiff nest]

yeah yeah my bad

how about

the range

[simple gazelle]

plug in f(g(24)) [ignore the restrictions here]

the range will be $(f(g(24)), \infty)$

[flat frigate]

MarcoMa210

[simple gazelle]

the range is basically where the output reaches

[safe radish]

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[pastel crane]

can someone help me solve for x

[timber karma]

67

[summer coral]

show what you have tried so far.

[pastel crane]

idk i js know x is 2017

[summer coral]

x is not 2017

that is way too high

[pastel crane]

but i tried it

[summer coral]

wait no that works

did a calculation wrong

[torn isle]

have u tried taking eeverything to one side?

[summer coral]

start by taking all the x's to one side and all the constants to the other side

[torn isle]

ye

[pastel crane]

ok i'll try

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[crisp swift]

how am i supposed to figure out whether if they r asking for bernoulli or exact differential question as they hv just given "solve" 😭😭

[fathom jewel]

hmm

You should know by identifying the equation

[crisp swift]

umm is there like a trick or smthng to it? 🥀

[fathom jewel]

Like the first one is obviouly more of a type of Bernoulli, right

the second one obviously more of a type of an exact differnetial eq

[crisp swift]

ohhhh yeahhhhhhhhhhhhhh umm can i ask one more doubt tho its regarding a question which i was solving rn but got confused lil

[fathom jewel]

sure

[crisp swift]

over here in the first one they didnt divide y by anything rather they jsut ignored it but in the 2nd one they didnt rather they divided but whyy

nd in few places after making rhs function of x
the leftover y values in the lhs they took the substitution method nd solved

like which is the correct method then??

[fathom jewel]

So by Bernoulli, you'd substitute t=1/y so I guess they tried to bring everything in the form of (1/y)

but it isn't necessary and division by y doesn't really change the methods

[crisp swift]

wait isnt both bernoulli?

[crisp swift]

soo we can ignore the y like we did in first right?

[fathom jewel]

The left photo looks like an exact one

they are also talking about an integrating factor

[crisp swift]

ohhhhhhhh

btw thenkss for the helpp
(this sub aint for me frfr)

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[fathom jewel]

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[lean otter]

any quick method

<@&286206848099549185>

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lean otter]

!helpers

[safe radish]

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

[plucky elk]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lean otter]

kk

[lean otter]

solve it

[summer coral]

no. helpers dont solve stuff they guide you

[lean otter]

i want quick method

for that type of question

[quiet plume]

Do you know what the eccentricity of an ellipse is?

[lean otter]

yeah

for standard equations b^2 = a^2(1-e^2) if a > b

and change a and b for b > a

[quiet plume]

You would need to find a and b then, which you can do by writing the equation in the standard form of an ellipse.

[near sky]

oppenheimer saying reiman to solve meth 🥀

[plucky elk]

Don't be entitled

[lean otter]

??

[plucky elk]

!vol

[safe radish]

Helpers are just people volunteering their time to help you. Be polite and patient.

[lean otter]

yeah i learned stand equations only

[lean otter]

uh huh ,,,,patient

i am not patient

but i am polite

[plucky elk]

This isn't polite nor patient

[summer coral]

well guess its time to learn 😭

[plucky elk]

Then go pay for a tutor

[lean otter]

sir pls tell me any quick method for that pls sir pls ...for this i shall always obligef to you pls sir

[lean otter]

already paying mann

[visual linden]

Bruh

[plucky elk]

I don't see money hitting my bank account

[lean otter]

i am not ill nor i have any disease then why i am patient

brooo

[pale dock]

Too bad lol

[summer coral]

like mentioned above helpers aint getting paid theyre choosing to help

[pale dock]

This isn't the place to ask a question that you need to be solved quickly, Ill say that

[visual linden]

yk this repels helpers from helping you right, if you're being disrespectful?

[summer coral]

so either suck it up and wait for someone to help or go ask ai for a quick wrong answer

[lean otter]

ai is jst shit mann

[summer coral]

yes i am aware of this

[lean otter]

t_t ig i have to think myself now

uh i am sleepy

[pale dock]

Yes which is why it's probably better to wait a few mins to get help

[lean otter]

i have to get the quick method otherwise i am goona cooked in tommorow classrom

[quiet plume]

The quick method is writing it in standard form.

Which you do by sending everything to one side, expanding (y-4)^2 and completing the square appropriately.

[lean otter]

can;t able to complete the square like we got 8/9 y^2

term

[quiet plume]

You can complete the square.

Just write $\frac{8}{9}y^2 - 8y = \frac{8}{9}\left(y^2-9y\right)$

[lean otter]

uh

[flat frigate]

Azyrashacorki

[lean otter]

u have to maintain 1 in rhs tooo

[plucky elk]

This isn't funny

[lean otter]

iam not saying it is funny

[plucky elk]

Then you should have been more patient and polite

[lean otter]

ban me

[plucky elk]

No one believes this besides you

[polar lynx]

I can mute or ban if you want?

[lean otter]

my english is weak dude

[polar lynx]

Seems like it would be better for you to quit bickering though and focus on the math.

[lean otter]

ban then bcz i got the method

ban me

[polar lynx]

Okie dokie

.close

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[near sky]

no way lad got banned

https://tenor.com/view/pepe-why-pepe-pepehands-crying-gif-12683546

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[heady grove]

I need a little help with an exercise that in Nielsen and Chuang, its about adjoints. I'm trying to read through Axlers Linear algebra done right but I feel like the notation in Nielsen and Chuang is throwing me off. Here is the exercise, it uses dirac notation so anyone who is familiar with it would be greatly appreciated

[heady grove]

I have attached only the definition that I am using which is the same as that in Axlers book

Here is the exercise

here $|w \rangle$ is a vector and $\langle v |$ is the dual.

[flat frigate]

Luke

[heady grove]

What I have done so far is change the $|w \rangle \langle v |$ into a linear operator

[flat frigate]

Luke

[heady grove]

I'm just wondering if there is a way to solve this problem by just using the definiton that is given at the start of the thread I made or do I have to use the fact that every linear operation can be written as a matrix representation and then somehow show that the RHS is true. I stopped working it out when I say "taking the adjoint" because I don't think that with the definition that is given that it is fair to just say that. I know that the adjoint is just the transpose of the complex conjugate but the definition that I am given here doesn't state that.

Does this suffice basically?

Except I'm not convinced by the swapping of indices just by the definition that is given

And that's my problem, if there is someone out there that is more familiar with linear algebra like this then I'd appreciate the help.
Thanks

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@heady grove Has your question been resolved?

[safe radish]

@heady grove Has your question been resolved?

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[astral gyro]

I need help with these 3 math problems they are Euler circuits i’ve been out of math because i had to be in the hospital i’m confused on these particular questions because anything i even put it says it’s wrong even if i beleive it’s “DNE”

[astral gyro]

i understand 0 odd vertices are a euler circuit and also a path and 2 odd vertices is a euler path but not a circcuit

<@&286206848099549185>

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[astral gyro]

my bad

[sinful hollow]

0odd vertices → euler circuit
2 odd vertices →euler path only
more than 2 →none........that quite simply is the golden rule

[astral gyro]

i guess i’m struggling to like formulate it onto the graph?? idk if that makes sense..

like for example question 7 you had to figure out what path that uses one doorway once i did what i thought it was and it wasn’t correct i did a bunch of other ways

idk i’m just confused on that pathway part stuff?

[sinful hollow]

0 odd vertices → there is an eulerian cycle (starting and ending at the same vertex)
2 odd vertices → There is only an euler path (starting at one odd vertex and ending at the other)
more than 2 → here is neither a path nor a cycle

[astral gyro]

yes i understand that part, just how to apply it to the graphs in the photos above?

to the questions that were asked

sorry if i am confusing n

!

[sinful hollow]

ok np 1 min pls

[astral gyro]

take your time!

[sinful hollow]

Well actually its not that difficult each room or space represents a vertex and each door is a side connecting two of them... all u have to do is calculate the angle of each vertex then apply the golden rule I told u about n then choose the start n end points

[astral gyro]

can you send an image representation of that? sorry i am a visual learner

[sinful hollow]

Oh sorry I'm really busy right now maybe later

[astral gyro]

no worries

[sinful hollow]

Someone else can help you with this

[astral gyro]

do i @ another helper?

[sinful hollow]

mmm maybe

[astral gyro]

okay!

<@&286206848099549185>

[solar hazel]

which one are you working on?

[astral gyro]

right now the house floor plan one!

[solar hazel]

okay

do you know what the graph looks like?

[solar hazel]

yea but like in terms of just vertices {A,B,C,D,E,F} and edges

[astral gyro]

sorry i don’t understand what you mean like
A has 2 vertices?

[solar hazel]

do you see like in question 9 here, there are points labelled A,B,C,D,E and they have some lines connecting to each other?

[astral gyro]

yes

[solar hazel]

in question 7, we could draw something like that, where each doorway between each room is like a line that connects those two rooms

we call those lines edges in the graph

the ‘points’ are the vertices in the graph

[astral gyro]

yes i understand that

[solar hazel]

it would look like that

[astral gyro]

OHHH OK

SORRY visualization helps a lot

[solar hazel]

a path through that uses every edge once in the graph is called an eulerian path

[astral gyro]

okay i see

i thought it was asking me something different

[solar hazel]

you’ve probably been given a theorem somewhere on when a graph has an eulerian path

[ionic schooner]

pardon me, but doesn't that need to also end at the same point?

[solar hazel]

i don’t understand. an eulerian cycle does that, but the question doesn’t ask for that as far as i can tell

[astral gyro]

i thought we were supposed to see what letters you go through and you couldn’t repeat / go over the same one

idk if that makes sense

so lmk

because it’s asking what path it takes

and what letters of the path it is

[solar hazel]

you can repeat letters

you can’t repeat doorways (edges)

[astral gyro]

ohhh okay

[solar hazel]

and you need to use every doorway once

you can try to just experiment until you find a path

[astral gyro]

okay that makes more sense

thank you

for expplaining

[solar hazel]

np

[safe radish]

@astral gyro Has your question been resolved?

[safe radish]

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[safe radish]

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[solar hazel]

<@&268886789983436800>

[clever tinsel]

<@&268886789983436800>

oops, sorry.

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[safe radish]

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[grand widget]

Hi

[opaque fern]

<@&268886789983436800>

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[winter whale]

server is being welcomed with lovely servers a lot today...

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[fringe drift]

im not sure what to do here

[fringe drift]

i can find the first derivative

[fathom jewel]

Differentiate both sides

[fringe drift]

yh thats gives 1/1+x^2

[fathom jewel]

Yes

Notice it's a geometric series

So you can say both sides have the same derivative

Now you only need to check if the antiderivatives differ by a constant

[fringe drift]

how do we know its

a geometric series

[fringe drift]

i dont get i

[fathom jewel]

So if you differentiate both sides you get a geometric series but you have -x² plugged in

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@fringe drift Has your question been resolved?

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[olive atlas]

I have the right angled triangle ABD, inside of which there is AMN.
Then i have the two trapezoids:

MOPN and OBDP.
For which:

Are aligned: A, M, O, B and A, N, P, D
MN // OP // BD

And eveything is right angled on A.
So, AM = AN, AO = AP, AB = AD

Etc.

We know that: AB = 12 and AM = 4sqrt(6)

I need to get the length of MO so that MOPN and OBDP are equal.

So I got the formula:

area MOPN = area OBDP
<=>
(AO×AP)/2 - (AM×AN)/2 = (AB×AD)/2 - (AO×AP)/2

But, from before we know that:

AO = AP = AM + MO

So we get the formula:

AM²+4AM×MO+2MO²=AB²

All fine till here

We define x,y,z such that:

x = MO, y = AM, z = AB

And we put the formula into its quadrstic form:
(-2)x + (-4y)x + (z²-y²)

And then I solve it and find: -2sqrt(30)-4sqrt(6)

Which isn't correct at all.
I don't understand where I made a mistake, I made one but I don't know where.

This is the scheme we had.
I think I overthought a lot during the process.
But I'd really like to know where I made a mistake (or maybe more).

Thank you a lot in advance.
Feel free to ping me for more precisions

[olive atlas]

I don't even think I need any quadratic formula.

[warm warren]

What is the question?

[olive atlas]

Find the length of the segment MO in order for MOPN and OBDP to be equal.

That was the question asked which I tried to answer.

[vague phoenix]

Do you have an original copy of the question by any chance?

[olive atlas]

It is exactly as I wrote, I did the translation, the original one is in french

[near sky]

[olive atlas]

So my questoom woukd be:

Where did I make one or several mistakes?

[vague phoenix]

Where is O located? On MN?

[olive atlas]

Between M and B as stated

[olive atlas]

Are aligned: A, M, O, B on one side and on the other: A, N, P, D

[vague phoenix]

Ah i see, the paragraph so long

[olive atlas]

Yeah, my apologies for that ^^'

I tried to be as precise as I could while giving ecerything I had

[near sky]

this ?

[vague phoenix]

I think you got the quadratic wrong

I got: $2x^2+4xy+y^2-z^2=0$

[flat frigate]

Minλ

[vague phoenix]

My bad i am on bed so the handwriting ugly ☕

@olive atlas my quadratic different from yours

[olive atlas]

Yeah you multiplied by -1?

[vague phoenix]

Why?

[olive atlas]

Or maybe I did indeed get the quadratic wrong

I think I did

[vague phoenix]

Yours is -2x-4yx+(z^2-y^2)

[olive atlas]

If you muktiply mine by -1 it gives yours no?

[vague phoenix]

No

Your x is power of 1, mine is power of 2

[olive atlas]

Oh yeah I js made typo

My bad

On ny paper I have written:

(-2)x²+(-4y)x+(z²-y²)

I made typo when I copied it in my paragraph ^^'

I'm very sorry for that ^^'

[vague phoenix]

Ah then its correct

Hmm this have 2 solutions

[olive atlas]

Yep!

[vague phoenix]

Did you take the negative solution or the positive one?

[olive atlas]

Then maybe it's the other one?

[olive atlas]

The negative one 😂

[vague phoenix]

Yes

[olive atlas]

That may explain it

Aaah damn I was correct but took the wrong answer... 🙃

But I think there is a way to solve thay without quadratic equations because were not supposed to have learnt then yet

[vague phoenix]

Yeah

[olive atlas]

This is out of the question though

Thank you a lot!!

I'll check the pisitive one when I finish my meal!

[vague phoenix]

But your way in finding the similar ratio is correct tho

No problem!

[olive atlas]

I'm glad I found it 😆
I was afraid I overhought too much over this

[safe radish]

@olive atlas Has your question been resolved?

[olive atlas]

Well, I have to calculate the positive answer now.

But yes.

[safe radish]

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[drifting spindle]

can anyone help ?

[near sky]

bruh

https://youtu.be/DbDjaJhmqkE?si=RtUjgsvjSwUEoLWN

[drifting spindle]

i knew bro was gonna send this

thx tho

[near sky]

...

np

[drifting spindle]

.close

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[vague phoenix]

https://tenor.com/view/riddler-paul-dano-new-rockstars-does-he-know-gif-26833694

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[normal moss]

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[spiral saddle]

how to prove gcd(a^n, b^m) = 1 if gcd(a,b) = 1 when a,b,n,m in Z

[quiet plume]

Did you not do this the other day?

You can use prime factorization.

[spiral saddle]

is there a simpler route?

bezouts maybe

@quiet plume

[quiet plume]

I wouldn't say it's not simple.

[gaunt elk]

write a = gcd(a,b)*x and b = gcd(a,b)*y with x and y coprime.

[spiral saddle]

how does that work

[gaunt elk]

$a^n = gcd(a,b)^n *x^n$

[flat frigate]

Lin Xia

[spiral saddle]

why

how

[gaunt elk]

it is just multiplication

$(ab)^n = a^n * b^n$ for all $a,b \in R, n\in Z$

[flat frigate]

Lin Xia

[quiet plume]

Does this not just say that x=a and y=b or is this the point?

[visual kraken]

Prime factorization is honestly probably the easiest route

Like, gcd(a,b)=1 means they have no primes in common

Then what possible primes could a^n and b^m have in common?

[gaunt elk]

probably but you have to master it a little bit. But prime factorization is not required.

[gaunt elk]

Yes probably not the better start i agree

[visual kraken]

I guess, but this method seems like a lot more work for the sake of not writing down p_1 * p_2 * ... * p_k a bunch of times

[gaunt elk]

Sure let's just use prime factorization then

[safe radish]

@spiral saddle Has your question been resolved?

[gaunt elk]

@spiral saddle do you know how to do it ?

[spiral saddle]

I do not

if you can help me I would appreciate it

[quiet plume]

You wrote it down yesterday, it's the same argument

[spiral saddle]

I dont want to use prime factorization tho

is messy and there is better way

trust me

[quiet plume]

It's not messy it's like 3 lines

But ok

[gaunt elk]

you can do it without the whole factorization

Let p be a prime divisor of gcd(a^n,b^m) what can you say about p ?

[visual kraken]

If you're sure. But the prime factorization method really is just one more step past what I said up here

[quiet plume]

You can use Bezout and expand stuff using the binomial theorem, but that is what I would call messy.

[spiral saddle]

suppose d = gcd(a^n, b^m) where d is not 1

then it follows d | a^n => d | a
similarly d | b^m => d | b

[visual kraken]

d | a^n doesn't imply d | a tho

[gaunt elk]

not true unless d is prime

[spiral saddle]

care to elaborate?

[visual kraken]

Take a = 6

a^2 = 36

4 | 36

But 4 | 6 is false

[spiral saddle]

help me

@visual kraken @gaunt elk @quiet plume

[gaunt elk]

@spiral saddle

[spiral saddle]

is just what I said

but d = p

[gaunt elk]

this is not the same thing

[spiral saddle]

basically if gcd is non 1 then there exists some prime that divides this gcd

[gaunt elk]

yes

so let's take p be such a number

[spiral saddle]

because then the prime factorization is filled with something

[spiral saddle]

p | a^n and p | a^m

what's your point?

[gaunt elk]

p is prime if and only if p|ab implies p|a or p|b.

[spiral saddle]

yeah euclid lemma, what about it?

[gaunt elk]

use it

here

[spiral saddle]

oh I see, is successive multiplication of p | a

basically p | a or p | a^(n-1)

[gaunt elk]

yes

[spiral saddle]

yeah if p is prime then p | a^n <=> p | a

[gaunt elk]

yes

[spiral saddle]

similarly p | b^m <=> p | b

therefore we get that p | gcd(a,b)

and since gcd(a,b) = 1

we need that p | 1

and there is no prime that divides that shit @gaunt elk

basically we get an absurdity

[gaunt elk]

yes so there is a contradiction

[spiral saddle]

@visual kraken

@quiet plume

is this proof ok?

[spiral saddle]

are you in highschool kid?

[gaunt elk]

not really

[spiral saddle]

you are strong for a highschooler

[spiral saddle]

college? workforce?

@gaunt elk

[gaunt elk]

lost

[spiral saddle]

i appreciate it dawg

.solved

[safe radish]

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[ebon grove]

I have this problem and part a) has me confused. It would seem that the first equation, b/(B-A), is the one I should use, and b would equal 0. But then I would get 0 for my probability, but not if I have P(X<-1) or something similar, which does not make sense given this is a uniform distribution. Should I be using the third equation instead?

[tardy mango]

the first equation is wrong

[light shoal]

that formula for P(X <= b) is not correct

[tardy mango]

should be b-A in the numerator

[light shoal]

also needs to be piecewise

[ebon grove]

lovely.

what is the correct formula, then?

[ebon grove]

that's what i thought

[light shoal]

(b-A) / (B-A) if A <= b <= B

0 if b < A
1 if b > B

[ebon grove]

ty

which then gives the intuitive answer of P(X<0) here being 1/2, since it's half of a uniform distribution

[light shoal]

yes, for any interval symmetric around 0, like [-5,5] is here

[ebon grove]

does it matter if the probability is asking for X strictly great/less than vs. greater/less than or equal to?

he does not have any of this in our slides

[safe radish]

@ebon grove Has your question been resolved?

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[copper wraith]

is this question correct?

[still sentinel]

I think I can reduce it to B u C u C' = C u C'. Which is equal to the universe we are working in.

Which means it's not correct? I am not quite sure...

[copper wraith]

i see

thanks

.close

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[copper wraith]

a number is purely real if z=z'

[copper wraith]

z' is complex conjugate of a number

this statment is written in my book so i was thinking around this statment