help-19

[tepid acorn]

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[young helm]

hello

[young helm]

need help with this

[mystic saffron]

sure, i'd suggest expressing 0.25 as a fraction

[young helm]

i did that

[mystic saffron]

that'll be easier to visualise

[young helm]

isolved itfurhter

[mystic saffron]

can you show your work?

[young helm]

yep gimme a sec

Same thing

Idk what to do further

hmmm

[calm raft]

my hint is $\frac{log_x(a)}{log_x(b)} = log_x(\frac{a}{b})$

[clever fjord]

lebesgue

[young helm]

oh

[calm raft]

wait

I think im wrong

[young helm]

yea a guy told me a differnt formula

[wild rune]

$log_{x}(a)/log_{x}(b)=log_{b}(a)$

[calm raft]

its wrong sorry I confused it w something else

[clever fjord]

The Great D

[calm raft]

yes that

[young helm]

well thats cool but what to do next after this formula

what to doafter expressingthem aspowers of 2

[calm raft]

then you simplify it?

[young helm]

how?

[calm raft]

its quite straightforward

[wild rune]

also $log(1/a)=-log(a)$

[clever fjord]

The Great D

[young helm]

oh

I want to know what to do after

I have never dealt with powers of base

Are their any rules for them

[wild rune]

Well you dont need to have a power in the base

[young helm]

Hmm

[wild rune]

reconsider the whole problem with 0.25 as a fraction

$1/a$

[clever fjord]

The Great D

[young helm]

Why does the question say to express them as powers of 2 then

[young helm]

Ok so it will become into 4?

[wild rune]

yes right track

[calm raft]

$\frac{log_5(2^{3})}{log_5(2^{-2})} = \frac{3}{-2} \frac{log_5(2)}{log_5(2)}$

[clever fjord]

lebesgue

[wild rune]

that

[young helm]

Ohhh

[young helm]

This makes so much sense

[calm raft]

yoru werukamu

[young helm]

So they final answer will be 3/-2 ?

[mystic saffron]

8 = 2^3

[young helm]

Yayyqya

[calm raft]

I miss highschool

I wish I could be a highschooler again, I'd dominate everyone fr

[mystic saffron]

same bruv

and then now you have to do lebesgue integrals?!?!

no offence to ur username 😉

[young helm]

Daym youd be a math bully

Thanks guys

You guys awesome

This group keeps me sane while studying maths

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[calm raft]

exactly

😪

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[thorn cloak]

Hia
I want to know how many bits (base 2 digits) of precision I need for something like sound analysis, for example if I can use float (53bits), or more loose approximation for something like sine cosine etc.

[low locust]

for what usage? not like nowadays you need to be careful with using too many bits

[thorn cloak]

Sorry what do you need to know in mode details?

[low locust]

I mean why are you worried about whether 53 might already be too much

[thorn cloak]

Also yea, it doesn't need to be exact, you may tell it roughly

[calm raft]

sound analysis? I've never heard of that before

can you enlighten me

[thorn cloak]

Like extracting info from sounds
And using that to generate (synthesis) sounds that we want, for example

[low locust]

just use floats or doubles. it doesnt matter

[odd edge]

@thorn cloak Has your question been resolved?

[odd edge]

@thorn cloak Has your question been resolved?

[thorn cloak]

Maybe can you make an high and low limit?

[low locust]

this just screams of premature optimization. on which toaster is your program supposed to run where you have to worry about which data type to use for numbers

[thorn cloak]

Is it for optimizations of my program?

Sorry if I said that
That sounds like a total misspeaking of me

Btw just curious: why do languages have different number types if it's premature optimizations to discuss which to use or whatever I did

[low locust]

well float vs double is a question of precision. but worrying about whether a float already uses too many bits is a question of premature optimisation

cause it almost certainly doesnt matter

but float vs double also doesnt matter basically always

[odd edge]

@thorn cloak Has your question been resolved?

[thorn cloak]

Ok so what about the other question? That being how accurate the sine and cosine approximations should be
And if it's impossible to answer exactly, you may answer roughly or possibly provide upper and lower bounds, thanks

[low locust]

well I dont know what you need them for. but I cant imagine that using a float would not be accurate enough

so just use that

dont worry too much

[thorn cloak]

🫡

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[warped bloom]

could i get a hint on how to solve this?

[warped bloom]

im really not sure on what to do

[fallow wolf]

Induction

[warped bloom]

Ohh

But how do I represent the odd divisors of n+1

[fallow wolf]

why would you need that when inducting?

oops you do need that, sorry

you can think about the largest power of 2 dividing n+1

say n+1=2^k*b where b is odd

by inductive hypothesis b+some other numbers =n^2

in the step, notice that 2n+1 is odd itself and 2n+2's largest odd divisor is b

so some other numbers+2n+1+b=n^2-b+2n+1+b=(n+1)^2

qed

@warped bloom

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@warped bloom Has your question been resolved?

[warped bloom]

Ah I see

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@warped bloom Has your question been resolved?

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[zealous skiff]

is this correct?

[odd edge]

@zealous skiff Has your question been resolved?

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[gleaming hedge]

need help researching if these series converge or diverge

[gleaming hedge]

this is what I've done so far using D'alambert:

but I don't want to be too quick and jump to conclusions

there's one condition in this problem:

this has the limit of e

but Idk what/where it's supposed to be used

[toxic monolith]

A few days ago I told you that this series does not respond to d'Alembert's criterion, but to Raabe's or Sitrling's formula, have you forgotten?

[gleaming hedge]

it's not the same one

but either way, could you check my work?

oh wait

[toxic monolith]

it is thwe same

[gleaming hedge]

it is

[toxic monolith]

it i salso in mybook

[gleaming hedge]

but why use Raabe since I think I get somewhere with D'alambert

[toxic monolith]

dAlemebrt is too weak for this series

becasue

it gets 1

and DAlembert does not help in it

[gleaming hedge]

how does it get 1 from e(n+1)/n?

[toxic monolith]

you wroly wrote

wronlgy*

[gleaming hedge]

how so?

[toxic monolith]

wait few minutes i shw you why

[gleaming hedge]

ok

[toxic monolith]

$\lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n \to \infty}\left[ \frac{e^{n+1}\left( n+1 \right)!}{\left( n+1 \right)^{n+1}}\cdot \frac{n^{n}}{n!e^{n}} \right]=\\=\lim_{n \to \infty}\frac{e^{n+1}\cdot n!\left( n+1 \right)\cdot n^{n}}{\left( n+1 \right)^{n}\left( n+1 \right)n!\cdot e^{n}}=\\=\lim_{n \to \infty}\left[ e\left( \frac{n}{n+1} \right)^{n} \right]=e\cdot \frac{1}{e}=1$

[clever fjord]

Joanna Angel

[gleaming hedge]

wait how

[toxic monolith]

I think I wrote everything correctly and precisely

[gleaming hedge]

I guess so?!

[toxic monolith]

do you have any doubts?

[gleaming hedge]

I have no clue wtf you're on about

whatsoever

[toxic monolith]

my students at uni, never have any doubts ) wanna ask them ? )

[gleaming hedge]

I think you're schizophrenically confused

all I asked is how you got that

not that you're wrong

e^n+1 is e^n*e?

so I have to be left with

e(n/n+1)^n

[toxic monolith]

that is true

[gleaming hedge]

ohhhhh

i see where I was wrong

thanks, me

[toxic monolith]

ok)

[gleaming hedge]

I'll close this after writing up something

.close

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[gleaming hedge]

for these series

[gleaming hedge]

using Raabe's criterion I get:

all good so far?

[toxic monolith]

yes it is )

[gleaming hedge]

so

[gleaming hedge]

so far so good, yes?

[toxic monolith]

yes all is ok so far

[gleaming hedge]

how do I diff this?

e is a constant

so it should be e^2 ? at the denom that is

[toxic monolith]

1/e is a constant, so take it out of the expression, and only differentiate numerator

[gleaming hedge]

ohh

[toxic monolith]

and to differentiate the nuemrator you shud apply such formula

[gleaming hedge]

it's (cx)'=c*(x)'

[toxic monolith]

a^b = exp [ blna]

nto c', but c

ah ok, i di dnot see it

[toxic monolith]

$\frac{1}{e}\left[ \left( 1+x \right)^{\frac{1}{x}} \right]'$

[gleaming hedge]

so far so good?

[clever fjord]

Joanna Angel

[toxic monolith]

yes

[gleaming hedge]

so then

[gleaming hedge]

that's true, right?

what do I do here?

[toxic monolith]

no

[gleaming hedge]

what do I do then?

[toxic monolith]

$\left{ f\left( x \right)^{g\left( x \right)} \right}'=\left{ f\left( x \right)^{g\left( x \right)} \right}\cdot \left[ g'\left( x \right)\ln\text{}f\left( x \right)+g\left( x \right)\cdot \frac{f'\left( x \right)}{f\left( x \right)} \right]$

[clever fjord]

Joanna Angel

[gleaming hedge]

like this?

[toxic monolith]

too much

middle term

is to eb removed

[gleaming hedge]

right

[gleaming hedge]

it's this?

[toxic monolith]

no

[toxic monolith]

$\left{ f\left( x \right)^{g\left( x \right)} \right}'=\left{ f\left( x \right)^{g\left( x \right)} \right}\cdot \left[ g'\left( x \right)\ln\text{}f\left( x \right)+g\left( x \right)\cdot \frac{f'\left( x \right)}{f\left( x \right)} \right]$

[clever fjord]

Joanna Angel

[gleaming hedge]

so it should be

[toxic monolith]

use this one i show you twice, that is special formula for functions called

power - exponential

[gleaming hedge]

yeah?

[toxic monolith]

$\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \left( \frac{1}{x} \right)'\ln\left( 1+x \right)+\frac{1}{x}\cdot \left[ \ln\left( 1+x \right) \right]' \right]$

[clever fjord]

Joanna Angel

[toxic monolith]

so yes

but do not forget

about first terms

[gleaming hedge]

right so

I get

that's true?

[toxic monolith]

right part apporaches 1, but do not set it as 1 , before you leave limit symbol

and

( 1 / x) ' = - 1 / x^2

[gleaming hedge]

do I not have to only work with the expression inside the brackets?

[toxic monolith]

you differentiate the expression inside, but rest must be written too

[gleaming hedge]

why do I differentiate it?

[toxic monolith]

i did few steps ahead in my mind, , you have to connect thsoe fractions together and you get 0/0

and you use Hospitale

twice even

[gleaming hedge]

w-what

which fractions?

this is a candidate for hopital?

[toxic monolith]

let me show oyu , min plz

[gleaming hedge]

oh ok sorry

[toxic monolith]

$\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \left( \frac{1}{x} \right)'\ln\left( 1+x \right)+\frac{1}{x}\cdot \left[ \ln\left( 1+x \right) \right]' \right]=\\=\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \frac{1}{x\left( 1+x \right)}-\frac{\ln\left( 1+x \right)}{x^{2}} \right]$

[clever fjord]

Joanna Angel

[toxic monolith]

and now

you have to subtract those fracions

connect them together

[gleaming hedge]

it's this/

I forgot how to find common denominator

[toxic monolith]

$\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \left( \frac{1}{x} \right)'\ln\left( 1+x \right)+\frac{1}{x}\cdot \left[ \ln\left( 1+x \right) \right]' \right]=\\=\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \frac{1}{x\left( 1+x \right)}-\frac{\ln\left( 1+x \right)}{x^{2}} \right]=\\=\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \frac{x-\left( 1+x \right)\ln\left( 1+x \right)}{x^{2}+x^{3}} \right]=$

[clever fjord]

Joanna Angel

[fallow wolf]

multiply top and bottom by something that makes the denominators same, right?

[gleaming hedge]

oh yeah

[fallow wolf]

just curious, what is this for?

[toxic monolith]

ilacertae computes ths limit to sue Raabe test , udner my supervision

[fallow wolf]

o

[gleaming hedge]

now I have

if I take x out of numerator and denominator, it should work out fine from here, right?

[toxic monolith]

youmissed (1+x) next to logarithm too

[gleaming hedge]

uhh

[toxic monolith]

$\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \left( \frac{1}{x} \right)'\ln\left( 1+x \right)+\frac{1}{x}\cdot \left[ \ln\left( 1+x \right) \right]' \right]=\\=\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \frac{1}{x\left( 1+x \right)}-\frac{\ln\left( 1+x \right)}{x^{2}} \right]=\\=\frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \left[ \frac{x-\left( 1+x \right)\ln\left( 1+x \right)}{x^{2}+x^{3}} \right]=$

[clever fjord]

Joanna Angel

[gleaming hedge]

no?

[toxic monolith]

almsot, but minus must be in front of third term

if you write minus, then 1 - 1 = 0 and you only get - ln(1+x)

[gleaming hedge]

yeah?

[toxic monolith]

1 - 1= 0

not -2

[gleaming hedge]

so then I was right from the start?!

so it's -ln(1+x)/(3x^2+2x) ?

[toxic monolith]

yes

and use H last time

finakll yi shwo you all solution how i did too btu plz continue

[gleaming hedge]

so for -1/2 the series is divergent

yes?

[toxic monolith]

yes )

look:

$\lim_{n \to\infty}\left[\frac{1}{e}\left( \frac{n+1}{n} \right)^{n}-1 \right]n=\left[ 0\cdot \infty\right]=\\=\lim_{n\to\infty}\frac{\frac{1}{e}\left( \frac{n+1}{n} \right)^{n}-1}{\frac{1}{n}}=\lim_{x \to 0^{+}}\frac{\frac{1}{e}\left( 1+x \right)^{\frac{1}{x}}-1}{x}=\\=\lim_{x \to 0^{+}}\left{ \frac{1}{e}\cdot \left( 1+x \right)^{\frac{1}{x}}\cdot \frac{d}{dx}\left[ \frac{\ln\left( 1+x \right)}{x} \right] \right}=\\=\frac{1}{e}\lim_{x \to 0^{+}}\left{ \left( 1+x \right)^{\frac{1}{x}}\cdot \frac{\frac{x}{1+x}-\ln\left( 1+x \right)}{x^{2}} \right}=\\=\frac{1}{e}\lim_{x \to 0^{+}}\left{ \left( 1+x \right)^{\frac{1}{x}}\cdot \frac{x-\left( 1+x \right)\ln\left( 1+x \right)}{x^{2}\left( 1+x \right)} \right}=\\=\frac{1}{e}\lim_{x \to 0^{+}}\left{ \left( 1+x \right)^{\frac{1}{x}}\cdot \frac{1-\ln\left( 1+x \right)-1}{2x+3x^{2}} \right}=_{\cdots }=\\=\frac{1}{e}\cdot e\cdot \left( -\frac{1}{2} \right)=-\frac{1}{2}$

[clever fjord]

Joanna Angel

[toxic monolith]

so Raabe defeated this series

if you used stirling then it wud be litle bit faster

though the limit wil look also ugly

[gleaming hedge]

We haven't studied Stirling

[toxic monolith]

i recommend it too

read abou tit

[gleaming hedge]

they will absolutely write me off if I do so

use it

[toxic monolith]

i see

[gleaming hedge]

they are absolutely serious if someone uses material we haven't studied

they literally write Fs

[toxic monolith]

🙂 ok

wel true, if a student can prve it then it is allowed

prove*

but you can read it just for your knowledge

Wallis formula and Stirlign formula

it is very nice part of analysis

[gleaming hedge]

tysm

gonna get to work on this:

[toxic monolith]

ok

🙂

[gleaming hedge]

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[tidal scarab]

So with cards and combinations, say i had a situation where 3 cards are dealt, then a card is burned and 2 cards are dealt to the player.

I believe there would be 52C3 combinations for the dealer's cards, but then for the player, how would I look at the situation? Would I be saying 51C2 since one card was burnt or 48C2 accounting for the already dealt cards?

[wooden python]

the burn card can be safely not given a shit about

cause you don't know what it is anyway

[meager juniper]

Well, if the cards are numbered 1-52, and the burned card is always the highest card, then you'd never get dealt the hand 51, 52, but provided there isn't a strategy to burning the cards, then 51C2 possibilities. And if there is a non-random strategy then that is 51C2 - 1 possibilities, but with a possibly non-uniform distribution?

[wooden python]

burning means taking a card off the deck face down

[meager juniper]

ah, I was not familiar with that term

[tidal scarab]

oh really

in that case would the player be seeing 52C2 combinations for their 2 cards or 49C2

since 3 were already dealt

[meager juniper]

52C2

[tidal scarab]

or is it 52C2 because they havent seen the dealer's cards yet

ok i see

[meager juniper]

So a good way to think about this is: let's ignore the dealer for now, and imagine we have a deck of 52 cards.

We deal 2 cards to the player, this is obviously 52C2 possibilities yeah?

What if we deal 2 cards from the bottom of the deck instead? Still 52C2.

What if we select two cards at random from anywhere in the deck, still 52C2.

Burning a card is equivalent to just putting one card on the bottom of the deck, and then dealing the top two cards.

Similarly, dealing three cards to the dealer, then burning a card is equivalent to putting 4 cards on the bottom of the deck then dealing two.

[tidal scarab]

oh ok

i see, thank you

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[mystic saffron]

his in three different ways and verify these all yield the same result. Additionally, interpret this result.

Hi, anybody here who can solve this problem?

I can't get the vector right

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

Can anybody help me please?

.close

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[thick pine]

first for c

will it be

x+a = -a? and x+a = a?

[odd edge]

@thick pine Has your question been resolved?

[thick pine]

<@&286206848099549185>

[ionic mulch]

You have to split (c) in 4 cases
When you explicit an abs value, you have to take it's argument positive and negative
For example, $|2x+a|$ will be 2x+a when 2x+a is greater than 0 or -2x-a when it is < 0

First case: 2x+a= x-a (assuming both modules are positive). The result will be x= -2a
Second case: 2x+a= -x-a(first is positive and second is negative). The result will be 3x=-2a or $x= (-3/2) *a$

[clever fjord]

Marian_is_here

[ionic mulch]

Will continue...

[thick pine]

oh, i thought we had to take all abs values to one side and the other values on the opposite

so we can solve it as is?

[ionic mulch]

You have 4 cases
Two when explicited abs values have the same sign
and another two when explicited abs values have different signs
Solve the equation and see the results(some results may be identical in some exercises)

[thick pine]

HAHA

im sorry i dont get it

can you do this q for me?

[ionic mulch]

I try to solve
I will show the whole solution to see it

[clever fjord]

Marian_is_here

[ionic mulch]

I post the answer too early, sorry

[thick pine]

wait what?

[thick pine]

yes i get this

[ionic mulch]

I pressed enter by mistake(when I don't want)

[thick pine]

oh lmao dw ty

[ionic mulch]

You want an explanation for (d) now?
I do not really understand what you want

You solve (d) with the same algorithm
Share the equation in 4 cases based on abs value explicitation
Solve the equation and see the results

[thick pine]

OKAY I'LL TRY

[odd edge]

@thick pine Has your question been resolved?

[odd edge]

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[thick pine]

.reopen

[odd edge]

✅

[thick pine]

<@&286206848099549185>

where do I even begin???

[quasi sparrow]

Quadratic formula

[thick pine]

without calc

LIKE HOW DO I PROVE

pmi? induction? contradiction? none of those?

[quasi sparrow]

The quadratic formula isn't calculus

[thick pine]

well i meant calculator

[quasi sparrow]

????

[thick pine]

but like

isnt it one of these?

[thick pine]

okay no i realized we can get the values without calculator

[quasi sparrow]

You can prove it in many ways

[thick pine]

but when it says prove, dont we have to use like another method?

[thick pine]

what are the ways?

[quasi sparrow]

Quadratic formula is simplest

[thick pine]

can you like jus list them?

[quasi sparrow]

No that'd be a waste of my time

[thick pine]

CMON

LIKE THE WAYS TO PROVE IT?

[quasi sparrow]

WHY DO YOU KEEP SHOUTING

[thick pine]

i apologize

sincerely

[signal oar]

For quadratics, the discriminant b^2 - 4ac shows how many real roots there are. For b^2 - 4ac > 0, there are two, for b^2 - 4ac = 0, there is one, for b^2 - 4ac < 0, there are none.

This is equivalent to what Riemann is saying, using the quadratic formula

[thick pine]

wait what i dont get it

what I ended up doing was i input it into the -b+root b^2-4ac/2a

[signal oar]

$x_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

[thick pine]

okay if its 2 real roots it'll be Discriminant > 0 right

so do we do

oh we do that only?

[signal oar]

See the b^2 - 4ac in there

[thick pine]

yup

[signal oar]

That's the discriminant

[thick pine]

sweet okay

so it'll be b^2 - 4ac > 0?

[signal oar]

If that's > 0, we'll have + some root and - some root

So two solutions

If it's 0, then it'll be +0 and -0

So one solution

If it's < 0, then there are no real solutions

[signal oar]

Yes

[thick pine]

i dont get how we prove it though? like wait okay so if i input it

4-4(a)(-7)

so 32

32 > 0

what now??

[signal oar]

... thus there are two real solutions

[thick pine]

THAT'S IT?

okay now what if i did

oh wait

THANK YOU

[signal oar]

You could also use Bolzano's theorem and such, but using the discriminant should really be the fastest way here

[thick pine]

psht okay no ill stikc with discriminant

[thick pine]

what does for all non-zero real values of k mean??

is it still the same thing?

b^2-4ac>0

kx^2-3x-k

[toxic monolith]

if k=0, yoru equaiton becomes a linear one, then it wud have only one solution

hence, k cant be a zero

[thick pine]

ooo okok

okay so will it be

-3^2 - 4(k)(-k)

wait what

9 + 4k^2 >0

[toxic monolith]

yes)

that is all

[thick pine]

oh sweet

[toxic monolith]

smiles

[thick pine]

ION

ION KNOW HOW TO DO THIS

dont even know where to begin

<@&286206848099549185>

[green elm]

what have you tried for (a)?

[thick pine]

can you like explain how we go for one of them

and i can like try to do that for everything else :)))

because i don't even know the first step :))))

[green elm]

ok, do you know that one form of a quadratic equation is:
y = a(x-b)(x-c)

that form will be useful for (a)

[thick pine]

okok

wait can you explain like the steps, as in what we have to do

[thick pine]

do we use the form for all of them? or just a?

[green elm]

well in the form y = a(x-b)(x-c), what's the value of y if x=b or x=c?

[green elm]

let's focus on (a) first

[thick pine]

y = a(b-b)(b-c)

[thick pine]

what is this form called?

[boreal crag]

simplify

[green elm]

factoring into linear factors i guess

[thick pine]

y = ab - ab ( b-c )

y = (b-c)?

[boreal crag]

no

[thick pine]

WHAT

[green elm]

one of your factors was (b-b)

which is better known as...?

[thick pine]

0?

[green elm]

yea

[thick pine]

oh so y=0

[green elm]

right

and same is true if x=c

[thick pine]

right

[green elm]

so b and c are the roots

what are the roots of the graph in (a)

[thick pine]

how do I tell?

2 and 6 are the x intercepts, 5 is the y intercept

[green elm]

roots are the same as x intercepts

they mean the same thing

[thick pine]

OH

2 and 6 then yes

[green elm]

the places where the y value is zero

yea, so b=2 and c=6, or vice versa

now you just need to find a

[thick pine]

y = a(x-b)(x-c)

so do I replace x with b or c?

[green elm]

no

you want to use that form
y = a(x-b)(x-c)
along with the other point you are given on the graph

the one at y=5

[thick pine]

oh so

5 = a(x-2)(x-6)

[green elm]

yea, but also plug in the x value of that point

y = 5 and x = what?

[thick pine]

x is 2 and 6 no?

oh wait is x =0

[green elm]

no, for the point on the graph

yes, x=0, y=5 is the given point

you know it has to satisfy the equation

so plug those values into y = a(x-2)(x-6)

then you can solve for a

[thick pine]

5 = a(-2)(-6)?

[green elm]

yep

[thick pine]

OHH

okay so

12a/5

what no

a=12/5

[green elm]

no...

[thick pine]

no

5/12 = a

[green elm]

yes

[thick pine]

YAY

and a is?

[green elm]

well you just said what it is

[thick pine]

oh

wait no what is it on the graph

[green elm]

oh i see

it's not always obvious what it is just by looking at the graph

[thick pine]

5=5/12(x-2)(x-6)

[green elm]

it's a scale factor that ensures the three points all lie on the graph, is all i can say

[thick pine]

oh, wll why did you use this formula?

[green elm]

should be y= on the left

[thick pine]

oh okok

[green elm]

because the graph for (a) tells you the two roots, so the formula involving the two roots is the natural one to use

as opposed to (b), which tells you the vertex and another point

so a different form will be more useful there

[thick pine]

OHH

how many formuls are there?

[green elm]

i would say there are three main ones

[thick pine]

now dis one, but it's only got one x intercept :((

[green elm]

y = a(x-b)(x-c)
y = d(x-e)^2 + f
y = gx^2 + hx + i

[thick pine]

what in the fuck

[green elm]

those are the three most useful forms

[thick pine]

duly noted

[green elm]

in each case you have three constants to find

i gave them different names for each equation to avoid confusion

[thick pine]

shweet

for this one though, we only have one root

so

[green elm]

yes

but it's also the vertex

[thick pine]

y=d(x-e)^2 + f?

[green elm]

yes you want that form

[thick pine]

FIGURED IT OUT

what's f

and what's d, and e

[green elm]

f is the y value of the vertex

e is the x value of the vertex

[thick pine]

omg ok

and d?

is there like a google img with all the constants, formulas, and etc

[odd edge]

@thick pine Has your question been resolved?

[odd edge]

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[scenic bramble]

i dont understand how the formula for the sum of geometric series up to n works.

if we want just one term, n, its a*r^n-1.

So it confuses me that somehow a(1-r^n)/(1-r) could magically produce a sum of all terms.

How could division by R give us a sum, a number bigger than just 1 term, if we're dividing it?

[midnight ferry]

I'll show you a quick and easy to undestand proof

Let $S = a + ar + ar^2 + ... + ar^n$

[clever fjord]

Eduardo

[dense violet]

I didn't understand how to solve it. I think there is a defect in the form, or I am the one who has a defect

[spiral basalt]

if 0 < r < 1, then 1-r < 1 so dividing by 1-r increases the quantity
So it's at least not nonsensical

[spiral basalt]

!occupied

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

[midnight ferry]

then $S\cdot r = ar + ar^2 + ar^3 + ... + ar^{n + 1}$

[clever fjord]

Eduardo

[midnight ferry]

this means that $S - Sr = a - a r^{n + 1}$

[spiral basalt]

if r > 1, I let you figure out why it's stilla greater quantity than the last term of the sum

[midnight ferry]

you can factor as: $S - Sr = S(1-r)$

[clever fjord]

Eduardo

[midnight ferry]

and if you devide by $1-r$ on both sides you get:

[clever fjord]

Eduardo

[midnight ferry]

$S = \frac{a-ar^{
n+1}}{1-r}$

[clever fjord]

Eduardo

[scenic bramble]

yea idk im just really slow on the uptake i guess, ive been trying to think about this for a while and cant wrap my head around it. I'll go through what you and eduardo have said and try again

[midnight ferry]

and this is the formula for the sum

[clever fjord]

Eduardo

[midnight ferry]

this was a correction of a typo here:

[midnight ferry]

.

This is the easyest way I know of proving this formula for the sum

If you don't undestand some step of the proof please let me know @scenic bramble

[spiral basalt]

induction is also pretty simple, but doesn't quite explain it like the explicitly telescopic proof

[low locust]

(telescoping arguments are theoretically induction, but its such an obvious induction that no one actually phrases it like that)

[scenic bramble]

thank you, ive seen this proof on youtube as well. The way you get to the proof makes sense, i just have no conceptual understanding for the formula and how it works when put into effect. Idk i think i need to just keep trying.

[midnight ferry]

Using the formula is very simple, just find what value of a and n you need and plug it in. Try to solve some exercises about geometric sequences to get a better understanding of this.

[scenic bramble]

right, i can use the formula, but i want to (maybe to a fault) understand why it works. For example, when i look at a*r^n-1 this is easy to visualize and intuitive, its taking 1 r away to account for the initial term ar^0. But the sum of partial terms formula i cant understand

well thank you all for the help. my inability to understand this after hours of thought and even having it explained to me, is very disheartening. but ill just keep at it.

.close

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[dense violet]

I didn't understand how to solve it. I think there is a defect in the form, or I am the one who has a defect

[odd edge]

@dense violet Has your question been resolved?

[modern hamlet]

What do you mean by "defect in the form"?

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[wanton fjord]

Given the complex numbers z and z' such that |z|=|z'|=1 and zz'+1 is different from 0, prove that w=(z+z')/(1+zz') is real

[wanton fjord]

I tried writing z=a+bi and z'=c+di and comparing w with its conjugate

[south plume]

z' = a-bi

by definition

(if z' is conjugate of z)

so try with z=a+bi and z'=a-bi

[wanton fjord]

in this case z' is not the conjugate

it's just another complex number

[south plume]

oh

what's your notation for conjugate?

*?

[wanton fjord]

mb we note the conjugate with a line above

[south plume]

ah okay

$w = \frac{a+bi+c+di}{1+(a+bi)(c+di)} = \frac{(a+c)+(b+d)i}{(1+ac-bd)+(cb+ad)i}$

[clever fjord]

artemetra

[south plume]

yeah you are probably not gonna get anywhere with that

have you tried converting it to polar form?

[wanton fjord]

No

Ill try

[tall veldt]

^was wondering why abs = 1 was important

[south plume]

yep

$z=e^{i\theta}, z_1=e^{i\theta_1}$

[clever fjord]

artemetra

[south plume]

$zz_1 + 1 \neq 0$\
$zz_1 \neq -1$\
$e^{i(\theta+\theta_1)} \neq 0$\
$\implies \theta+\theta_1 \neq \pi + 2\pi k, k\in\bZ$

[clever fjord]

artemetra

[south plume]

i guess

[south plume]

also $w \in \bR \iff k(z+z')=(1+zz')$ for some $k \in \bR$ OR $((z+z'),(1+zz') \in \bR)$

[clever fjord]

artemetra

[odd edge]

@wanton fjord Has your question been resolved?

[wanton fjord]

thanks

.close

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[rustic geyser]

hello how do i integrate this

with x and y being in the same thing

ive not come across this before id appreciate some guidance on what to do next

[haughty barn]

use chain rule, $$dy/dx = dy/dt * dt/dx$$

[clever fjord]

🅷ศຮຮศས

[haughty barn]

differentiate y(t) to get dy/dt and x(t) to get dx/dt

[leaden karma]

they have done the first part already.

[rustic geyser]

yeah

im asking how to finish my integral

[haughty barn]

ah mb

[leaden karma]

to integrate, the integrand needs to be a function of x only.

there should be no y terms

try to write the y^2 in terms of x

[rustic geyser]

oh ok

so

sqrt1-x^2

well

[leaden karma]

well, $|y|=\sqrt{1-x^2}$ yes

[rustic geyser]

1-x^2

[clever fjord]

kheerii

[leaden karma]

what is y^2 then

[rustic geyser]

ye

[leaden karma]

right

[rustic geyser]

Ok

lemme try

[leaden karma]

u can use that to solve the integral

[rustic geyser]

thanks

[haughty barn]

isnt this just the formulae for arc length ?

[rustic geyser]

idk

[leaden karma]

that's what the question is asking OP to calculate..

[rustic geyser]

its a length of a circle

so yeah

[leaden karma]

yeah, you can do that directly if you see that the parametric functions define a unit circle

also, in the first part, you need to define dy/dx as a function of x only

[rustic geyser]

oh ye

thanks

[leaden karma]

so replace y with $\pm\sqrt{1-x^2}$ (it's a bit ambiguous since y is not really a function of x)

[clever fjord]

kheerii

[leaden karma]

y can be both positive and negative

[rustic geyser]

okay

lemme try this integral

oh i got the formula for length wrong lol

i did -

[leaden karma]

yeah I just saw that

[rustic geyser]

well this is easy

just arcsinx

thanks for the help

[leaden karma]

yea

[rustic geyser]

.close

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[bleak obsidian]

Does anyone understand how to solve this question, so far from what I know that the probability of 1st box is 1/8. But i got confuse where the 2nd box's probability is.

[bleak obsidian]

But im guessing that it might be 7/8 for box 2 since Monty opens box 3 to 8 which are empty, just confirming if my answer is correct.

<@&286206848099549185>

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@bleak obsidian Has your question been resolved?

[sleek moth]

I think so yeah

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[bleak obsidian]

Thanks

.close

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[mystic saffron]

how do u denote this in conventional sigma notation?

[mystic saffron]

[
\sum_{k=-m}^n a_k r^k
] im guessing?

[iron bear]

laurent series

looks good to me

[mystic saffron]

also like

[
\sum_{-\infty}^\infty
]

[mystic saffron]

how tf do u write something like this

[iron bear]

thats how you write it

[mystic saffron]

that feels so weird tho

[quasi sparrow]

n in Z

[mystic saffron]

[iron bear]

[ \sum_{k = -\infty}^\infty = \sum_{k \in \Z} ]

[mystic saffron]

thats even weirder

[mystic saffron]

yeah i will do the k in z

probs better

ok

thanks

,close

,close

.close

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[iron bear]

thank riemann too

[quasi sparrow]

.coose

[mystic saffron]

that was directed at both of u smh

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[mystic saffron]

At this year's big book sale, the price of a book with math illustrations was reduced by 40%. Math teacher Jimmy wants the book but waits until the end of the sale period when it is sold at 50% off the sale price. Jimmy calculates that he got the book SEK 126 cheaper compared to the regular price. How much did the book cost without the sale?

[mystic saffron]

I don’t know how my equation should look like

What do they mean with regular price

[analog basin]

not including the sale price

ie not reducing it by 40%

[analog basin]

is 180 the answer?

[mystic saffron]

Yes

How

[analog basin]

alr

so let x be the original price

[mystic saffron]

Mhm

[analog basin]

so when price is reduced by 40% the new price is $x - \frac{40}{100}x$

[clever fjord]

Wither

[mystic saffron]

X0,6

[analog basin]

$x - \frac{40}{100}x$

[clever fjord]

Wither

[mystic saffron]

Oh

[analog basin]

this will be the new price

[pearl patio]

where did you get the 40%?

[analog basin]

which will be $\frac{3x}{5}$ on further simplification

[clever fjord]

Wither

[analog basin]

sale price reduced by 40%

[pearl patio]

yeah and jimmy didn't buy it then

[mystic saffron]

Yeah

[pearl patio]

he bought it when it dipped to 50%

[analog basin]

yup

[mystic saffron]

Can’t I just write it as 0,6x

[analog basin]

u can

[mystic saffron]

Cuz it’s ab factor change or whatever it’s called

[analog basin]

yea

[mystic saffron]

Ok what’s next

[analog basin]

so whe price is 50%

it will be 0.3x

[mystic saffron]

Wait why

[analog basin]

50% means 1/2

[mystic saffron]

So why not 0,5x

[analog basin]

0.5 x 0.6

because you had the 0.6 already

[mystic saffron]

Ohhh ok

[pearl patio]

did you translate this question btw?

[analog basin]

yes

[mystic saffron]

Yeah

[pearl patio]

Seems ambiguous lol

Oh yeah that's why

[analog basin]

hmm

[plain badge]

180 is answer?

[mystic saffron]

Yes

But what’s the next step

[plain badge]

i just solved it mentally

[analog basin]

ok then finally equate it to x - 126

[plain badge]

guys -_-

[analog basin]

because the price off by 126

so 0.3x = x - 126

@mystic saffron

[mystic saffron]

Ok ok wait I’m calculating it

[analog basin]

alr

[mystic saffron]

Uhhh

I got a long decimal number

[analog basin]

how?

did you solve this?

0.3x = x - 126

[mystic saffron]

😭

[analog basin]

what

it will be -0.7x not -1.3x

[mystic saffron]

Ohhh

[analog basin]

now do it

[plain badge]

Don't use calculator

it will ruin your brain

no cap

[mystic saffron]

It already has

[plain badge]

STOP using it

[plain badge]

it will hurt but yeah

the early u discontinue it

[mystic saffron]

I didn’t use it and I got it incorrect lol so I need to use it

[analog basin]

nice

[mystic saffron]

Thank youu

[analog basin]

no probs

[plain badge]

NOOOO

[mystic saffron]

How else like

[plain badge]

Do it again

try doing it with pen and paper

[mystic saffron]

I haven’t learnt how do divide big numbers

And other stuff

[plain badge]

long division?

[mystic saffron]

Why not pen and iPad

[plain badge]

whatever works

use the pencil and the ipad

[mystic saffron]

Yeah

[plain badge]

just refrain using calc

[mystic saffron]

I’ve tried

It has always failed me so

.close

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[versed obsidian]

hello i was trying to solve this differential equation but idk if my solution is correct

[versed obsidian]

$y'+y^{2}=x$

[clever fjord]

Jill ♡

[versed obsidian]

this is my solution: $\tanh\left(x\sqrt{x}\right)\sqrt{x}$

[clever fjord]

Jill ♡

[lean willow]

,w y'+y^2=x

[mystic saffron]

this is not linear

[lean willow]

I forgot how to do non linear ODEs or I didn't learn it

[mystic saffron]

won't have a nice elementary solution

you need some Bessel functions shit

[versed obsidian]

my solutions seems to be somewhat right though

[versed obsidian]

not every non linear requires special functions

y'+y^2=1 for example has a nice solution

[quasi sparrow]

Just plug it in and check them

[versed obsidian]

i just plotted it

[lean willow]

I think because we can write it as
$$\int \frac{dy}{1-y^2}= \int dx$$

[clever fjord]

Fractalogist

[quasi sparrow]

Why can't you plug it into your DE

[lean willow]

,w plot tanh(xsqrt(x))sqrt(x)

[clever fjord]

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[lean willow]

,w derivative of tanh(xsqrt(x))sqrt(x)

[clever fjord]

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[lean willow]

,w ((tanh(xsqrt(x))sqrt(x))^2

[clever fjord]

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[versed obsidian]

something is wrong im pretty sure

[lean willow]

Eh dunno if it would be equal to x

[versed obsidian]

looks like this

[versed obsidian]

tanh(x) not tan(x)

[quasi sparrow]

How are these two equal

[tall veldt]

oop yeah

so clearly not a solution

[versed obsidian]

theyre not thats why im asking lol

[lean willow]

Would be a good approximation tho

[quasi sparrow]

Wot

[mystic saffron]

i dont understand what you're asking for tbh

[versed obsidian]

^

[mystic saffron]

The solution is very not elementary according to Wolframalpha

[quasi sparrow]

So you answered your own question

[lean willow]

It seems we can't solve it

If the solution looks so compex

[versed obsidian]

yea but whats the correct solution

[mystic saffron]

its not elementary

[quasi sparrow]

.

[tall veldt]

,w solve y' + y^2 = x

[clever fjord]

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[versed obsidian]

yea this is not helpful

[versed obsidian]

whats the representation then?

in terms of other functions

[mystic saffron]

this

[tall veldt]

but this is the solution

not every DE solution can be expressed in normal functions

[lean willow]

,w bessel function

[clever fjord]

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[brittle beacon]

Are you supposed to actually solve the DE?

[versed obsidian]

why are there complex numbers though? does it have complex inputs?

[quasi sparrow]

Oh we're getting xy'd

[mystic saffron]

!xy

[odd edge]

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.

[versed obsidian]

no i just made it up myself

but apparently its hard lol

guess ill take the complex solution

[lean willow]

Solve y'+y''=x instead lol

[versed obsidian]

no. screw 2nd orders..

[lean willow]

It's 1000x easier than that

[versed obsidian]

all the ways to solve it suck

[mystic saffron]

if you are practicing ODEs just dont do nonlinear ones that are just crazy shit. Do the ones where you can get nice series solutions or something

[versed obsidian]

well you cant really see if an equation has a nice solution or not

[mystic saffron]

not coming up with ones on your own and instead following a textbook or a problem set solves that

[versed obsidian]

give me an easier one please i dont have a textbook on hand

[mystic saffron]

ok what have you studied so far?

in terms of ODEs

[versed obsidian]

mainly first order

nothing 2nd order yet

[quasi sparrow]

#book-recommendations

[versed obsidian]

yea okay but its gonna take a while to order something

you got anything for right now?

[mystic saffron]

[
(4+t^2)y' + + 2ty = 4t
]

[mystic saffron]

solve this

[versed obsidian]

also i already have euclid's elements to read beforehand

[versed obsidian]

thanks

:)

[quasi sparrow]

https://tutorial.math.lamar.edu/classes/de/de.aspx

[versed obsidian]

.close

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[sleek coyote]

can't you use u sub ?

its pretty simple and efficient, you can solve that one with like 3 lines.

for this specific integral its much better

well you can always try u sub, it would be clear if it doesn't work or not, but in most times if you know Integration by parts and other integration techniques you would see which is the best one to use.

well first assign a value for u and du

du is the derivative of u so whatever you assign u you have to take the derivative of it

in this specific example make 1+x^5 = u

can you tell me what du is ?

yes but you have to add dx,
so its du = 5x^4 dx

when you are using u sub, you are trying to make the integral simpler by subtituing u, but when if you substitute now you will have (integral) of (x^5/u) * dx

you have to remove dx so you can integrate du

[sleek coyote]

how do you make dx the only one on the right side?

correct

now put everything in the integral

see what happens

thats integration by parts, IBP and not u sub

but yes it can used to solve those kind of problems

did you get an answer to your integral?

after putting everything in, it becomes like this no ?

yes, when you have the value for dx and u

yes

what can you do here to simplify more

yeah

the x^4 goes with the x^4

since you are left with a constant you put it outside the integral

you are left with 1/u

remember that the integral of (f'(x)/f(x)) is ln(f(x))

well yeah but remove u

get back the 1+x^5

and dont forget the 1/5

check it out on youtube and google, a lot of people have done explanations on it.

well you have the correct final answer

i've never heard of your 2 methods up there

the second one looks like a harder u sub

but doesn't seem correct

actually the first one seems to have a correct final answer, but ive never used it and it seems jumbled

no problem.

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[unborn osprey]

Struggling with the start of q12, I tried to form equations but they just didn’t line up at all

[unborn osprey]

Nvm they meet at the origin,…..

Ignore this

.close

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[wind viper]

Hi

[wind viper]

need some guidance
what can i do now
cuz of the x idk how to get it to i(n-1)
and i dont want to look at the solution yet

[odd edge]

@wind viper Has your question been resolved?

[mystic saffron]

It's asking you to use integration by parts, so you express $(1-x^2)^n$ as $1 \cdot (1-x^2)^n$ and so integrate 1 and derive the $(1-x^2)^n$

[clever fjord]

Palahoo

[bold marten]

he knows

[mystic saffron]

Oh, so you actually got that $I_n (x) = 2n \int_0^1 x^2 (1-x^2)^{n-1} dx$

[clever fjord]

Palahoo

[mystic saffron]

Thanks!

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[bold marten]

@wind viper

https://math.stackexchange.com/questions/314602/derive-recursion-formula-for-an-integral

[mystic saffron]

You could try to transform it into $2n \int_0^1 (1-x^2) (1-x^2)^{n-1}$

[bold marten]

look what i found :3

[clever fjord]

Palahoo

[mystic saffron]

Oh, that's the answer

[bold marten]

pointless

it doesnt help you since they are still factors

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[wet lichen]

can someone look over this and see if i have the right idea?

[odd edge]

@wet lichen Has your question been resolved?

[dim thicket]

yes thats good

[wet lichen]

so a differential equation is trying to find how the rate of change of a variable changes in relation to the variable? Just to fully get it

this is the most confusing unit for me

[dim thicket]

in that case yes in general not really

[wet lichen]

what would it be more generally

[dim thicket]

differential equations can involve second derivatives and higher order

[wet lichen]

for first order differential equations would this work across the board?

[dim thicket]

no

integrating works for separable differential equations