help-0

[half epoch]

The inside of that is 1 comparison + 1 assignment(maybe). Now would you mind if I said that was 1 + maybe 1 = 1.5?

[alpine sable]

ok

[half epoch]

It’s not really important how long it actually takes, it’s constant time, which you’ll see soon

So that loop we get $\sum_{j=i+1}^{n} 1.5$

[ocean seal]

Learath2

[half epoch]

Putting it all together we have $\sum_{i=1}^{n-1} (1 + \sum_{j=i+1}^{n} (1.5) + 1)$

[ocean seal]

Learath2

[half epoch]

Yep. Now let’s start using some summation rules to simplify this

[alpine sable]

yes lets

[half epoch]

First one I’d like to use is $\sum_{i=1}^{n} c = nc \text{ where } c \in \mathbb{R}$

[ocean seal]

Learath2

[half epoch]

I think this one is simple enough, no?

We are adding together $n$ $c$’s. Which is nothing but multiplication

[ocean seal]

Learath2

[alpine sable]

oh i guess i've been evaluating these wrong. i thought summations were inherently in the form 1+2+3...+n

aren't they? otherwise why wouldn't one just use multiplication in the first place

[half epoch]

They aren’t. That’s $\sum_{i=1}^{n} i$ e.g.

[ocean seal]

Learath2

[alpine sable]

what is the i=1 in the above statement

[half epoch]

This is exactly how the authors handled that left summation after the split. If the inside is a constant, it’s nothing but multiplication

[alpine sable]

this one

you could see where the variables being in different places is confusing me

[half epoch]

i is called the index of summation, 1 is the lower bound

[alpine sable]

meaning it goes down to one or you need at least 1 number or what

[half epoch]

We start with i = 1. At each step we add 1 to i while it’s still smaller or equal to n

Think of it like a for loop.

for(int i = 1; i <= n; i++)

[alpine sable]

then what does this i represent if we aren't incrementing

[half epoch]

We increment i, but our term we are summing doesn’t involve i. So it doesn’t change

We take n steps until i = n. And at each step we add a c to the sum

[alpine sable]

ok i guess i understand that. its such a confusing way of writing things

[half epoch]

Yep

[alpine sable]

ok ok

[half epoch]

Ok, how about you use that rule to get rid of the summation we have for the inner loop?

[alpine sable]

which i can see is 25, 5*5

well i don't understand how the j=i+1 would increment or what n is so but i guess it'd be 1.5n

[half epoch]

Well we start at 1, we go until n. That’s n iterations. If we instead start at i+1 then we have i less iterations to make

n iterations - i iterations = ?

[alpine sable]

ok i see now

the summation of n - i iterations

[half epoch]

Yep, so after that simplification what do we have for the total time complexity?

Merge the +1s too just to clean it up more

[alpine sable]

ok i guess i don't see, i don't know why it wouldn't instead be n + i

[half epoch]

Instead of i and n, put random numbers there and it might make more sense

Going from 1 to 10, we’d take 10 steps, if we start at 5 instead, we’d take 4 less, no?

[alpine sable]

9 steps

if we're at 1 it takes 9 steps to get to 10

[half epoch]

Ah, that’s the problem with the steps analogy. We do evaluate the sum at the upper bound too, so that’s 1 extra term

As in it takes 9 steps, but there are 10 terms to our sum

[alpine sable]

right right i sort of lost track in whats going on, we're evaluating the overall summation term which right now stands at 1+ 1.5n +1?

[half epoch]

It’s not 1.5n. If j started at 1 it would be. But we are starting i later than 1

Instead of n terms, we’ll have (n-i) terms

Think of it with numbers if the symbols are confusing

[half epoch]

What’d happen if you added 1 to the lower bound? What if you added 2?

[half epoch]

So by adding 1, we have 1 less term to our summation

If we added 2 we’d have 2 less. If we add i we’ll have i less

[alpine sable]

i get that but yeah the j throws me

anyhow let me try to evaluate the summation in the inner part

[half epoch]

I’ll need to go for dinner in like 3 minutes :/

[alpine sable]

np

1.5(n-i)

[half epoch]

Put the 1s together aswell and we have $\sum 2 + 1.5(n-i)$

[ocean seal]

Learath2

[half epoch]

Now we use the sum properties to slowly split this apart. We can do $\sum 2 + \sum 1.5(n-i)$

[ocean seal]

Learath2

[half epoch]

You can also take a scalar factor out of a sum so $\sum 2 + 1.5 \sum (n-i)$

[ocean seal]

Learath2

[half epoch]

The only thing that remains is using the properties of landau symbols (O, o, big theta) to get rid of the constants. A scalar factor and a constant don’t change the time complexity.

$O(a + bn) = O(n)$ thus we arrive where the authors arrived

[ocean seal]

Learath2

[half epoch]

Thus they ignored constant stuff from the very beginning

[alpine sable]

the answer is theta(n^2)

i could see this being the justification for the inner loop being equal to theta(n)?

but i don't understand in their answer where the n^2 terms come from or anything past the first or second expression

the answer is here:

in fact this makes it seem as if the entire complexity comes down to the inner loop and nothing else

bc as you showed, the inner loop is equal to the n - i summation term

shouldn't it then be above the summation symbol? what makes it summation(n - i)

i really need to understand these summations bc they get much more complicated

[tacit arch]

how many iterations in the j loop on line 3?

[alpine sable]

i think they showed it to be n - i

iterations

[tacit arch]

so that's the inner loop

[alpine sable]

so they're just ignoring all the constants and linear runtimes associated with any other line and declaring the entire algo runtime equal to the inner loops complexity?

and furthermore how does that one summation on the left expand into all the terms and expressions following the equals sign

[tacit arch]

what all other constants and linear runtimes are you talking about

there's only one operation inside the j loop

[alpine sable]

i mean the rest of the algorithm that isn't the inner loop

[tacit arch]

have you taken calculus?

[alpine sable]

yes but i don't remember anything

[tacit arch]

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-3/a/review-summation-notation

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

i still don't get this

[tacit arch]

specifically what

[alpine sable]

in the videos it is obvious how to evaluate the summation. here it is not straightforward what they did or why

[tacit arch]

this is the value of the inner loop

[alpine sable]

forget about the pseudocode

[tacit arch]

and also the n-i here

[alpine sable]

i just want to know how to loop at that summation and evaluate it or expand as they did

right wouldn't that just become n-1 automatically

when i = 1

[tacit arch]

yes

but i ranges from 1 to n-1

[tacit arch]

you need to understand this to understand the solution

[alpine sable]

yes those were too easy

they don't have any variables

[tacit arch]

this n-1 comes from line 1 in the algorithm

[alpine sable]

again if we just forget the code for a second, i'd like to know if given that first expression, how it becomes n(n-1) - summation(i) which i understand to be the rule that you can break apart terms added or subtracted in a summation. ok fair enough. but to my eye it looks like they multiplied n*(n-1) and substituted in the summation of i term for i, which seems to be two different operations

[tacit arch]

oh

you can factor constants out of a sum

(n + n + n + ... + n) = n * (1 + 1 + 1 + ... + 1)

since there are (n-1) terms, adding 1 that many times is (n-1)

[tacit arch]

what do your arrows mean

use your words

[alpine sable]

it looks like they multiplied n*(n-1) and substituted in the summation of i term for i, which seems to be two different operations

[tacit arch]

did you follow this?

[tacit arch]

do not think of it like that. that's only true for starting index=1

[alpine sable]

i understand it when i plug numbers in yes. i cannot see what's going on when there are so many variables

[tacit arch]

?

[alpine sable]

thats what i'm referring to. yes i understand that 5+5+5+5 = 5*(1+1+1+1)

[tacit arch]

do you understand why
$\sum_{i=1}^{n-1} n = n(n-1)$ now?

[ocean seal]

riemann

[alpine sable]

no

[tacit arch]

can you simplify the last part here?

[alpine sable]

its a summation right

[tacit arch]

using this

[alpine sable]

summations are typically something + something different + something different, otherwise we would simply use multiplication

[tacit arch]

yes, typically

but not always

[alpine sable]

therefore, why is there any multiplication involved at all, when i evaluated the summations in the khan academy video it was k/2 + k/3 + k/4 + ... + k/n

[tacit arch]

so you need to understand how summations work generally

[tacit arch]

what's 1 + 1 + 1 + 1 = ?

[alpine sable]

4

[tacit arch]

observe that 4 = 5-1

so your example 5 + 5 + 5 + 5 = 5 * (5-1)

which is the formula in the screenshot for n = 5

[alpine sable]

what screenshot can you copy the link pls

[tacit arch]

$\sum_{i=1}^{n-1}n = n(n-1)$

[ocean seal]

riemann

[tacit arch]

the first part here

[alpine sable]

how do you know when i = n-1?

[tacit arch]

how do you know what?

[alpine sable]

nvm i just saw how to do it on paper

with 5 i = n-1 when i = 4

ok i think i'm getting it. the way math people think is wild. i would never look at any number added a bunch of times and think "why that's simply the number times the expression that number minus 1!"

anyway.. i guess that's how we get the first term, and then i isn't manipulated at all?

[alpine sable]

.close

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[cedar field]

How is du/2x a product?

[karmic rapids]

$\frac{1}{x} = x^{-1}$

[ocean seal]

illuminator3

[wary stream]

I'm pretty sure I explained to you that all the terms are getting multiplied together

[cedar field]

i know I didnt understand that one part very well and i didnt think i should dm you... can i ?

[wary stream]

No

[cedar field]

oh

ok

[tacit arch]

you should understand how multiplying fractions work

[wary stream]

It's literally what it means, all the terms are getting multiplied

Same idea as 2x(3c)4d

That's a product of terms

$2x(3c)\frac{3}{4d}$

[ocean seal]

dldh06

[wary stream]

That's a product of terms

[tacit arch]

watch this
https://www.youtube.com/watch?v=93Y8hCoOAj0

[lone heart]

@cedar field Has your question been resolved?

[cedar field]

thanks

everyone

.close

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[alpine sable]

.close

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[faint hamlet]

i dont get it

[spring patrol]

little tips for vectors, the length of your vector is the diff, between both X's and Y's (so Xnum2 - Xnum1 is = the X length)

[faint hamlet]

but why do we need the x length for?

[placid zinc]

You'll want to draw your vector as the hypotenuse of a triangle

Then, find the angle of said triangle

[faint hamlet]

ohhh

so i need to use soh cah toa

[placid zinc]

x-length is, of course, one of the triangle's legs

[faint hamlet]

.close

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[iron mulch]

!15min

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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.

[lone heart]

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[bleak cape]

Hello! I need help finding the value of m for which f(x) = -x^4 + 4x^3 + 8x^2 + m has two real solutions on the interval of [-1;-4]

[copper stone]

can you calculate the determinant ?

having 2 real solutions means that ∆ ≥ 0

[bleak cape]

yeah, in case of the equation being quadratic

it's x^4 though

[copper stone]

shit i didnt read that my bad

[bleak cape]

all good

<@&286206848099549185>

[iron mulch]

!15min

[lone heart]

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.

[bleak cape]

thanks for reminding me, you are really useful, if you paid attention you'd see it's been more than 10 minutes, no one answered my question

[rugged sun]

Hi again

[bleak cape]

hey

why is everyone telling me to get the discriminant 😔

[rugged sun]

Maybe there is a way to use it by doing a substitution first I am not sure yet

Just to clarify, when you solutions, do you mean roots?

[bleak cape]

yes

[rugged sun]

Ok

[bleak cape]

i know how to find the range of m for the roots on the graph, but i have no idea how to do it analytically

[rugged sun]

Ok, I will have a think

[iron mulch]

15 minutes is how long you must wait. Not ten. Maybe you can pay attention

[bleak cape]

i asked the same question on another thread, waited almost an hour for an appropiate answer, no one took hold, and I had to close the thread

[iron mulch]

Well that doesn’t matter. You still must follow rules even if you make a new thread

[bleak cape]

okay, i'm sorry

[lone heart]

@bleak cape Has your question been resolved?

[rugged sun]

Sorry for the delay, just a moment

[bleak cape]

No problem!

[rugged sun]

So I have a slight issue with this problem, are you sure the interval is [-4,-1]

[bleak cape]

-1 to -4

wait

yes

I'll send the full exercise in a sec

[rugged sun]

Yeah so that won't work imo

[bleak cape]

okay so, basically it says

let f(x) be 8x² + 4x³ - x⁴ + m

find out
a) local extreme points(i don't know if i translated it well, but it's the points where the function changes its growth(increases/decreases)

b) real values of m for which f(x) = 0 has two roots on the [a,b] segment, where x = a and x = b , a<b, are maximum local points of f(x)

i derived f(x), found that -4 and -1 are maximum points, and 0 is a minimum point

so we work with -4 and -1

[rugged sun]

The maxima are at x=-1 and x=4 I think

[bleak cape]

ohhh

it could be that

[rugged sun]

Well it is helpful to sketch the curve here and reason from that

[bleak cape]

yes, i did sketch it

a little

yes you're right it's 4 and -1

i messed up a sign

[rugged sun]

So as you move the curve up and down, between what values of m can you get two real (and I'm assuming distinct) roots

[bleak cape]

yes

that's what I did with the graph

but is there any way to find it out analytically?

without looking at the graph?

just mathematically

[rugged sun]

Honestly, I personally default to drawing the graph for these questions as the geometric intuition is faster than looking for an algebraic reason. I'm sure that there may be one, but honestly I don't see it rn, and its mechanism may very well mirror the geometric explanation anyway

Sorry for not offering a more concrete answer

[bleak cape]

Noo that's good

don't apologize

could you give me a tip on how to sketch a graph better?

or do i just take values from each range and put them on a graph

[rugged sun]

Well after you have plotted the coordinates of the maxima and minima, and you from the fact that the x^4 coefficient is negative that it must be going to negative infinity as x increases then all the roots kind of just place themselves

[bleak cape]

ohhh

[rugged sun]

When you link eveything up

[bleak cape]

i just figured out its gonna be pretty hard to understand a graph like this

[rugged sun]

The shape of this kind of graph can be deduced from knowing the coordinates of its stationary points and what type they are (max min inflexion etc)

[rugged sun]

Then you just pass a curve that has all of these features

[bleak cape]

but basically the line which intersects the graph is the solution right?

and the slider is the range

[rugged sun]

Yeah basjcally

I did it so the whole graph moves up and down, but it's thebsame thing

[bleak cape]

gonna be hard doing that without a calculator and on paper

like this?

[rugged sun]

But again, you can always calculate the derivative and determine he nature of the statioary points by hand

[bleak cape]

yes

[rugged sun]

Yeah

[bleak cape]

damn, wish me luck tomorrow

[rugged sun]

You have somehing?

[bleak cape]

yeah, i have an entrance exam

[rugged sun]

Oh wow, ok

[bleak cape]

these few exercises ive asked you to help are the first ones, 1 to 4, but the rest is probability which im good at

[rugged sun]

Well best of luck luck

[bleak cape]

but calculus, ehhh

[bleak cape]

Thank you!

[rugged sun]

Ok, well is there antthjng else I can help with

[bleak cape]

Nope (:

thank you so much

[rugged sun]

No problem at all, sorry if my explanation for the drawing was a bit iffy, i can elaborate there if you like

[bleak cape]

noo, i understood everything

thank you for the help, i really appreciate it

have a great day!

[rugged sun]

That's alright, good luck again

[bleak cape]

+10 social capital

[rugged sun]

You too!

Thanks

[bleak cape]

.close

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[thorny pulsar]

is sinx +1 the same as 1 +sinx

[thorny pulsar]

sorry quick question

[fierce prairie]

yes

it’s the same property as 4 + 6 = 6 + 4

[thorny pulsar]

second guessing myself thank you

over thought it

thank you

.close

[fierce prairie]

np

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[iron mulch]

Because it endlessly repeats so there’s no need to continue it?

Bro we didn’t make it we don’t know

They just chose to do so

Bro stop asking we don’t know they just did that

I’m literally a helper not the person who makes your assignments

We don’t know literslly we can’t answer this

They chose to do it like that there’s no specific reason

I’m literally a helper

Waste of time

.close

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[sly mantle]

lets chill

[iron mulch]

That’s a mod^

[real gazelle]

both of you chill

[iron mulch]

Enough already pls

[lone heart]

✅

[real gazelle]

the stopping at 6pi is arbitrary

[iron mulch]

I’m a volunteer can’t be fired

[night geyser]

you absolutely can be, fyi

[sly mantle]

@fathom sequoia last chance, pls stop harassing users

[real gazelle]

there's no reason it's 6pi and not 4pi or 8pi or whatever

[iron mulch]

i meant as I’m not like a job fired more like dismissed

[sly mantle]

@fathom sequoia come back in 24h

[iron mulch]

Please stop pinging me Tom 👍🏻

Oh

.close

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[oak storm]

https://cdn.discordapp.com/attachments/274757025341112320/1001302626899599420/Screen_Shot_2022-07-25_at_18.39.05.png

[oak storm]

this is kind of a silly question, but its been a long time and im not exactly sure how to factor it

i tried using online calculators but they gave me different answers, the correct answer should be F=25.62

[abstract fractal]

Why not replace the sines and cosines with their respective values

That oughtta clean things up a bit

[oak storm]

is there a way to do it without replacing numbers?

it makes it a bit messy

[abstract fractal]

I mean, yeah

Distribute out the 1/0.15

Move all the terms without an F to the other side

To some wacky factoring and solve for F

[oak storm]

this is what my professor did to solve it

but i wasnt sure his individual steps

[gray isle]

don't let the trig functions intimidate you

work towards
a * F = stuff

[oak storm]

i think im confused about what he moved to the other side

and how the sin 50 was divided

[alpine sable]

The terms which are multiplied by F

Then factor out F and divide other side by the factored out terms

[gray isle]

take it one step at a time

get terms with F on one side and terms without F on the other

this might be less intimidating
$$a(Fb-k)-c-Fd=0$$

[ocean seal]

ℝamonov

[oak storm]

okie ty, I’ll take a look once I’m home

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@oak storm Has your question been resolved?

[lone heart]

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[slender granite]

I'm trying to figure out the rules of dividing one term by another

[slender granite]

Like 2x^2 +3x+4/2x-7

[last ether]

Well

Long division

Or break up terms

[slender granite]

What's long division?@last ether

And for that matter what do you mean with breaking up terms?

[last ether]

Long division is like

Well you can Google that

It's stupid dumb to explain over text

[slender granite]

Ok

[naive hearth]

is it this?

https://www.mathsisfun.com/algebra/polynomials-division-long.html

[last ether]

Breaking up terms is like $4x \Rightarrow 3x + x$

[ocean seal]

Umbraleviathan

[last ether]

Which can make division easier as a fractional visual

[slender granite]

Ooh, you mean polynomial division that you use for polynomials of third degree?

[last ether]

Any polynomial

Any

4th degree, 69th degree, 62838328th degree

[slender granite]

And what use is breaking up terms?

Well, i kinda got it thx

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[cedar field]

@last ether ...

[last ether]

Yeah so

What's the original problem

@cedar field

[cedar field]

How is du/2x being multiplied?

[last ether]

Oh yeah

That's because dx is kinda like a factor of the integrand

So when we replace a factor with something, that value something also becomes a factor

What happens next is that (4x)/(2x) = 2

So it becomes $\int 2u^3du$

[ocean seal]

Umbraleviathan

[cedar field]

and what does this do?

[wary stream]

I don't understand what you're confused about this. It's literally a product, all the terms are getting multiplied together

[last ether]

You solve it

[cedar field]

how is it a product? thats my question?

[wary stream]

Because it's getting multiplied

That's how multiplication works

[cedar field]

where is the multiplication, how do I find it?

[wary stream]

You don't find multiplication, it's just multiplication

Like 2x(3)4x, those are getting multiplied, right?

[cedar field]

3 is being multiplied... it has brackets.

[wary stream]

So is the 2x and the 4x

Not all the terms needs brackets

Brackets just help make it easier to denote it's multiplication

2x(3)4x is the same as (2x)(3)(4x)

[cedar field]

so 2x(4)3x 5x 2x. would everything be being multiplied.

[wary stream]

I wouldn't suggest using spaces to denote multiplication, it doesn't make it that clear

Brackets or the multiplication sign is the best option

For fractions, you don't need to include brackets if it's clear enough, like $2x\frac{1}{4x}$

[ocean seal]

dldh06

[wary stream]

That looks like a product

And $\frac{1}{2x} \frac{1}{3x}$ is clear enough that it's a product

[ocean seal]

dldh06

[cedar field]

2x(3)4x so how would you be sure that its multiplication?

[wary stream]

What else would it be?

Have you not seen multiplication denoted like that before?

[cedar field]

addition, subtraction or division

[wary stream]

No

[cedar field]

why not

[wary stream]

Because there's no addition, subtraction, or division sign used

[cedar field]

oh

so what would it be for

2x(3)+5 4x?

[wary stream]

What exactly are you asking?

There isn't enough clarity for 5 4x

[cedar field]

what would you do with 4x

[wary stream]

As I mentioned, using spaces to denote multiplication doesn't provide clarity

Same idea as 2x, that term itself is a product with no brackets

It's 2 * x, is it not?

[cedar field]

like in 4x(u)^3 du/2x there is no addition, subtraction, multiplication and division between du/2x and u^3 like 5 between 4x?

[last ether]

Well there is division because you're multiplying by a fraction

[wary stream]

The division is clearly defined with the fraction bar

[last ether]

Because of multiplication properties, (4x)/(2x) -> 2

[cedar field]

there isnt a space in 4x(u)^3 du/2x

[last ether]

That's a fractional bar

[cedar field]

yes i know

[wary stream]

How would you explain that du/2x is getting multiplied with the rest of the terms?

[cedar field]

ye how?

[wary stream]

Because this was the 3rd I've explained it

It's literally a concept of multiplication

[last ether]

Here

$$4x(u)^3\frac{du}{2x}$$
$$4x \cdot u^3 \cdot du \cdot \frac{1}{2x}$$
$$4x \cdot \frac{1}{2x} \cdot u^3 \cdot du$$
$$\frac{4x}{2x}\cdot u^3 \cdot du$$
$$2\cdot u^3 du$$

Bruh

[wary stream]

Have you seen $2x$ being denoted as $2 * x$ or $(2)x$ or $2(x)$ or $2\times x$ or $(2)(x)$? They all mean the same thing

[ocean seal]

dldh06

[cedar field]

yes

ohhh

[ocean seal]

Umbraleviathan

[cedar field]

i might get it

thanks for all the time Umbraleviathan and dldh06

.close

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[versed root]

how do i calcualte the probibility of winning in rock paper sissers 10 times in a row

[proud stirrup]

Depends on the number of people u are playing with. If its just two people, then its 33%. Winning 10 times in a row is (1/3)^10

[versed root]

okay, thank you

!close

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[lone hamlet]

how do i do f

[lone hamlet]

ive underlined it

but im not sure

im using sine rule now

[proud stirrup]

shouldn't theta be 60 degrees? It says it is a regular pentagon for inner and outer.

[lone hamlet]

ohh no

(n-2)x180

[proud stirrup]

regular pentagon is composed of 5 equilateral triangles

[lone hamlet]

thats the formula for shapes angles

[proud stirrup]

oh mb

[lone hamlet]

and then u divide by 2

yeah

if u do that u get 54

and the triangle u get should be an iscoceles with 54, 54 and 72

now we need to the bottome bit

im using sine rule to find that

but im not sure how to implement it

ive forgotten how to

[vale wigeon]

no it isn't

[proud stirrup]

okay so you start by using sin rule

[proud stirrup]

yeah I got confused for a sec mb

[proud stirrup]

cos 54 = x/17.01

x is half the distance of AC

so after u find x times it by 2

[lone hamlet]

waiti i got the answer

i did sine rule

i got 17.01xsin(72)/sin(54)

can osmeone put this in the calc

my calc is broken

[proud stirrup]

this should be answer I think

[lone hamlet]

ohh thx

yep ima close this channel

thanos for the help

i was just extra dumb today

.close

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[cedar field]

@wary stream Really sorry for the ping but yesterday you said this... Why is it divided by 2?

[cedar field]

if someone else wants to help just tell me and Ill tell you background knowledge.

anyone?

[tacit arch]

after learning how to multiply fractions, you need to learn how to simplify fractions

[cedar field]

ok?

so can you help with this first?

[tacit arch]

that's what you need to do

cancel terms in the numerator and denominator

[cedar field]

ok

welll

thanks

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[normal ingot]

Given an IVP, $\frac{dy}{dy}=\frac{1}{(y+2)^2}$ and $y(0)=1$ state the domain of definition of the solution

[ocean seal]

lirmirit

[normal ingot]

so i managed to get the solution $y=-2+\sqrt[3]{3t+27}$ but since $y$ cannot equal -2, I got the restriction that $t\neq-9$.

[ocean seal]

lirmirit

[normal ingot]

however, the answer says that the domain is $t>-9$. why can't $t$ be less than -9?

[ocean seal]

lirmirit

[normal ingot]

(typo: the first msg should say $\frac{dy}{dt}$)

[ocean seal]

lirmirit

[tacit arch]

do you know the domain of $\sqrt[3]{x}$ ?

[ocean seal]

riemann

[normal ingot]

isnt that just all real numbers?

[tacit arch]

,w plot x^(1/3)

[ocean seal]

Results provided by WolframAlpha

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

Requested by riemann#2529

[normal ingot]

wait...

nvm

i see the mistake

thx

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[normal ingot]

.reopen

[lone heart]

✅

[normal ingot]

wait why is only the positive numbers considered to be the domain of cbrt

[tacit arch]

OH

wolfie defaulted to principal root when real-valued root is an option

[normal ingot]

bruh

[tacit arch]

,w solve dy/dx = 1/(y+2)^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by riemann#2529

[normal ingot]

yea thats what i got (with c=19)

[tacit arch]

alright no idea then. maybe something to do with t < -9 then y is multivalued

[normal ingot]

it seems like the solution is still defined when t is less than -9 :/

[tacit arch]

oh i know. the solution is not differentiable at t=-9

so you can't calculate $dy/dt$

[ocean seal]

riemann

[normal ingot]

as far as i know, the solution still differentiable at t=-10

[tacit arch]

yea but that branch doesn't connect to the y(0) = 1 branch

without the derivative blowing up to infinity

[normal ingot]

does the solution have to be continuous?

[tacit arch]

solutions to differential equations need to be differentiable

[normal ingot]

ah i see

that makes sense

thx

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[stable night]

hi can someone give me an idea on how to start on this

i think a direct proof would be hard for me

should i try contradiction?

[placid zinc]

I assume that [n/2] means round, and therefore it will always round up?

[stable night]

err i think its floor

round down

sry i didnt take the picture properly

[placid zinc]

Consider two different unconnected sections. Considering the degree that any random vertex must have, how many verticies must be in both sections, at least?

[ocean seal]

HellO

[stable night]

err im not sure

[placid zinc]

Some random vertex has to be connected to [n/2] others, so the section it is in must have at least [n/2]+1 verticies

[stable night]

wait why n/2 others?

[placid zinc]

Degree of any vertex is [n/2]

[stable night]

im sorry but why? cant the degree of a graph be 1

like if the graph has 50 vertices

there can be a vertex with a degree of 1

[placid zinc]

Not in this graph, by the question

[stable night]

oh

omg

ah okay

that makes sense

[stable night]

okay yes

then the maximum amount of edges that G can contain is n(n-1)/2

[placid zinc]

Let's say there's two different sections. Then this graph would have 2[n/2] + 2 verticies

[stable night]

yea

[placid zinc]

If n even, that's n+2. Too many.
If n odd, that's (n-1)+2 = n+1. Too many.

This graph has more verticies than it has. Contradiction.

[stable night]

ahh we had to account for even and odd

[placid zinc]

Well, no, but that's a pretty easy way to show that 2[n/2]+2 > n

I can't think of another atm haha

[stable night]

okay so the idea is:
to prove by contradiction ill show p and not q
which in this case is every vertex has [n/2] and the graph G is not connected
so we split the graph into two, and since there must exist a vertex in both graphs that has [n/2] vertices, the entire left/right subgraph needs to have at least [n/2] + 1 vertices

uh wait

vertices

from there we add the vertices from both graphs and we will get 2[n/2] + 2 which is greater than n

okay i get

[placid zinc]

@stable night
Well, pretty close. We can't simply assume there's two subgraphs

I don't think "subgraph" is the right word for this, but I don't know what the word should be haha

[stable night]

hmm

[stable night]

why tho?

[lone heart]

@stable night Has your question been resolved?

[placid zinc]

Would this suddenly be possible if there were 3?

I just threw out "let's pretend there are 2 sections"

Maybe something about 3 makes the contradiction go away.

[stable night]

wait why?

wouldnt there be even more vertices

[vale wigeon]

whats the original problem

[vale wigeon]

is it this?

[stable night]

yeah

[vale wigeon]

when you say section do you mean connected componnet

[stable night]

i understand the proof, its just that idk why cant we assume theres two subgraphs

[vale wigeon]

component*

do you mean 2 connected components

because (a) nobody said there can't be more than 2 connected components, and (b) this assumption is actually unnecessary

||let n = 2k + r where k ∈ N and r ∈ {0,1}, so that floor(n/2) = k

assume towards a contradiction that there exist two vertices u, v ∈ G with no path between them
in particular this means no vertex is a neighbor of both u and v at once (for if there were such a vertex x, we would have the path uxv go from u to v, which goes against our assumption)
both u and v have at least k neighbors each, as per the problem statement
but then the total number of vertices in the graph is at least k (u's neighbors) + k (v's neighbors) + 2 (u and v themselves) = 2k+2 > n
contradiction||

[stable night]

the n = 2k + r is for both odd and even cases rite

[vale wigeon]

yes

consolidating them into one until the need comes to split them (if at all)

[stable night]

ahh okay

i understand

thanks

[lone heart]

@stable night Has your question been resolved?

[lone heart]

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[quasi rover]

Try factorising, perhaps

[patent hedge]

wdym

factor the start?

[lilac nest]

Factor the first fraction

[patent hedge]

ok then what

.close

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[flat saddle]

Help, what is this, I don't understand it at all

[flat saddle]

I'm supposed to use this somehow

[lone heart]

@flat saddle Has your question been resolved?

[alpine sable]

mans learning enchantment language

[lone heart]

@flat saddle Has your question been resolved?

[flat saddle]

https://tenor.com/view/ah-shit-here-we-go-again-ah-shit-cj-gta-gta-san-andreas-gif-13933485

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[lyric grove]

What is meant by pulling out a factor a? We can assume a and L are positive and also wolfram alpha end ups with sqrt(a^2 + L^2) when you assume a and L are positive. (https://www.wolframalpha.com/input?i2d=true&i=a+*+sqrt\(40)Divide[Power[a%2C2]%2BPower[L%2C2]%2CPower[a%2C2]]\(41))

[lyric grove]

So my question is how would you go from a * sqrt(Divide[Power[a,2]+Power[L,2],Power[a,2]]) to sqrt(Power[a,2] + Power[L,2])

[gilded citrus]

multiply numerator and denominator by a

$a\sqrt{1 + (\frac{L}{a})^2} = \sqrt{a^2 + a^2(\frac{L}{a})^2} = \sqrt{a^2 + L^2}$

[ocean seal]

hsp

[lyric grove]

so a = sqrt(a^2) and you multiply both expressions with it.

Not sure why I haven't thought of this.. Thanks!!

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[keen python]

(n-1)! = n! * (n-1) ?

[worn fox]

plug in some numbers for n and see for yourself

[lone heart]

@keen python Has your question been resolved?

[keen python]

no doesnt look like

is there no way i can rewrite it to smth like n! * ... ?

[minor needle]

in this case it should be (n-1)! = (n-2)! * (n-1)

we generally write that property as n! = (n-1)! * n or (n+1)! = n! * (n+1)

[keen python]

i see thx

.close

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[delicate glen]

When I use the chain rule to differentiate, my derivative is different to the one on the mark scheme

[delicate glen]

i get

$\frac{dy}{dx}=\frac{5x^2}{2}\left(5x-1\right)^{\frac{-1}{2}}$

[ocean seal]

katie <3

[delicate glen]

mark scheme for reference

[pliant dune]

it looks like you've just differentiated the square root term and treated x^2 as a constant

[delicate glen]

i tried to do the dy/dx = dy/du * du/dx thing

$\frac{dy}{du}=\frac{x^2}{2}u^{\frac{-1}{2}}$

[ocean seal]

katie <3

[delicate glen]

$\frac{du}{dx}=5$

[ocean seal]

katie <3

[delicate glen]

where $u=5x-1$

[ocean seal]

katie <3

[wary stream]

You can't really apply u sub here. It doesn't work out nicely, I believe. You have to use product rule

[delicate glen]

BUT ITS IN A CHAIN RULE WORKSHEET

:(

[wary stream]

Then when you do chain rule, it just applied to the (5x - 1)^1/2. I don't think you did anything with the x^2

[pliant dune]

the problem here is you've assumed you can treat x as a constant when differentiating by u

[wary stream]

Can you post your full work for clarity?

[delicate glen]

sure if you can read my handwriting

[pliant dune]

for this sort of problem the easiest way is to use the product rule

then to use it in this case you'll need to use the chain rule similarly to what you did

[delicate glen]

ok ty

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[errant cedar]

I’ve got the parametric which is x=8t+4 y=5t+12 z=2t+15

But I’m not sure how to get to ii

Also the question is confusing to me haha like which line passes through that point

[worn fox]

They'll mean your new line passes through that point

You need to use your parametric equations to find the direction vector of your line

(Your first one)

[errant cedar]

How do I find the direction vector though?

[worn fox]

you can write the parametric equations in vector form

r(t) = at + b

a is the direction vector

[errant cedar]

Oh I see so direction vector is 8,5,2

Alright I know where to go from there

Thanks!

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[gilded vessel]

i generally perform well in all areas of math that i am taught but for some reason i just cant seem to grasp geometry and perform as well as i would hope to

[gilded vessel]

is there any possible reason for it or could anyone point me to some resource for help?

[languid bolt]

what part of geometry?

[wary stream]

Khan academy is a good website

[languid bolt]

Organic Chemistry Tutor is a good youtube channel

3b1b is also a good youtube channel, but idk if they have geometry

[wary stream]

That’s more for advanced maths, not geometry

[languid bolt]

oh ok

[gilded vessel]

i dont know just geometry in general

[languid bolt]

try Khan Academy ig

[lone heart]

@gilded vessel Has your question been resolved?

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[next flax]

could someone help me go through this basic rational equation question

i've been reviewing quadratic equations but before i do that i want to start with non-quadratic rationals

[chrome plank]

The problem is that x in the denominator, so a good idea might be to start by multiplying both sides by ||(x-2)||

[next flax]

yeah

[chrome plank]

So it simplifies and it's not a problem any more

[next flax]

uhh

i have now

5x-5 = (1/3)(x-2)

?

[chrome plank]

Yes

[vale wigeon]

yes, keep going

[next flax]

okay, and now i should get rid of the constant

(put 5 on the opposite side)

so 5x-5+5 = (1/3)(x-2) + 5

?

[vale wigeon]

sure, that's legal

[next flax]

so here's where i am stuck

5x = (1/3)(x-2)+5

the thing i am thinking to do

is divide both sides by 5

[vale wigeon]

legal but unhelpful

[next flax]

but that doesn't really give us a valid answer

for x

[chrome plank]

Bring every x term to the same side

[vale wigeon]

i would backtrack to here and multiply both sides by 3

[next flax]

that would make it a quadratic

[vale wigeon]

no, it would not.

[next flax]

if it equals 0?

[next flax]

ah okay lets try that

[vale wigeon]

do i understand correctly that your logic is "if the right-hand side of an equation is 0 then it is a quadratic equation"?

[next flax]

yeah

[vale wigeon]

this logic is incorrect

[next flax]

oh

[vale wigeon]

by that logic, the equation 7x - 4 = 0 would be quadratic

which it is not

[next flax]

well it equals zero

and has a squared power

whoops

[vale wigeon]

where

[next flax]

my bad, i forgot about that property

[vale wigeon]

where do you see a ^2 in 7x - 4 = 0

[next flax]

no dude

i'm saying a quadratic has a squared coefficient

[vale wigeon]

the name quadratic equation comes from latin quadratus which means square

[next flax]

that's what i am missing in my misconception

[vale wigeon]

what makes a quadratic equation quadratic is the presence of an x^2 term, NOT one of the sides being zero...

[next flax]

yes, allow me to correct myself 😂

[vale wigeon]

in any case

[chrome plank]

(also, to make it a quadratic you can't have any powers bigger than 2 to the x)

[vale wigeon]

from here i suggest multiplying both sides by 3

[next flax]

yeah

then it would get very messy

[next flax]

yeah so,

15x - 15 = (x - 2)

[vale wigeon]

yes, keep going

15x - 15 = x - 2

[next flax]

ah crap

i got a unhelpful answer again

add 2 to both sides side

and now i got x = 15x + 13

[vale wigeon]

there are still things you can do from here

[next flax]

like what

[vale wigeon]

well... first of all you should've gotten 15x - 13 = x

so you need to fix that sign error on your part

but aside from that

why not subtract x from both sides

[next flax]

whoops

yeah, my bad

[next flax]

because then x dissapears?

[vale wigeon]

what do you mean?

[next flax]

(-x + x - 2) = -x + 15x - 15

[vale wigeon]

.............

well, ok, you backtracked another step, but whatever

[next flax]

?

[vale wigeon]

yes, ok, so you have -x + x - 2 = -x + 15x - 15.

simplify on either side.

[next flax]

-2 = 15 - 15?

???

[next flax]

i'd rather go back to the place i was, i misinterpreted this as "do this instead of what you tried"

[vale wigeon]

-x + 15x is not 15.

if you have 15 apples and give away an apple, you aren't left with the platonic number fifteen in your fruit basket.

you're left with 14 apples.

[next flax]

okay

[vale wigeon]

and likewise, 15x - x is not 15, it is 14x.

[next flax]

ah i see whoops

[vale wigeon]

so you have 14x - 15 = -2.

[next flax]

yeahhhh

add 15 on both sides

14x = 13

x = 13/14

😄

[lone heart]

@next flax Has your question been resolved?

[vale wigeon]

i thought you solved your equation now

why press no on the bot's prompt

[next flax]

why does this person subtract the 4k to the right side instead of dividing it?

[vale wigeon]

what do you mean?

[next flax]

i divided 8k / 4k

and got 2k

he/she did 8k - 4k

[vale wigeon]

(8k)/(4k) is not 2k

[next flax]

fuck

[vale wigeon]

also keep in mind that you need to do things to BOTH sides

$\frac{4k-10}{4k} = \frac{8k}{4k}$ would've left you with a comparatively nasty fraction on the left.

[ocean seal]

Ann

[vale wigeon]

in effect it would almost be a step directly backwards compared to the previous

[next flax]

regardless

it's 2 or 2k

[vale wigeon]

while what you said is technically correct,

[next flax]

why did he/she decide to do 8k - 4k?

[vale wigeon]

(8k)/(4k) is 2.

subtracting 4k from both sides (which you should not describe as 'doing 8k - 4k') would leave you with no terms containing k on the left.

which is a good thing, because generally an equation with the unknown appearing only on one side is generally easier to solve than when the unknown appears on both sides.

[next flax]

but i just don't understand where i was supposed to make the inference that this was the RIGHT thing to do

if i did what i did

i got k = 5

when the right answer is 2.5

[vale wigeon]

i think you may have royally screwed up in a subsequent step.

it is impossible to tell where that happened without seeing all of your work.

or maybe, "what you did" was itself not valid at all.

if you can show your work then i can pick it apart.

[vale wigeon]

there's also the issue of you apparently thinking that there is only ever one and only one right thing to do.

[next flax]

I tried doing it another way but it still left me with 5

[vale wigeon]

alright i see the mistake

[next flax]

What is it

[vale wigeon]

you didn't divide both sides by 4k properly

specifically you didn't divide the left-hand side by 4k

$\frac{4k-10}{4k}$ is not the same as $\frac{4k}{4k} - 10$

[ocean seal]

Ann

[vale wigeon]

also there is another issue that is more notational in nature

[next flax]

but i did the latter ?

[vale wigeon]

and you were supposed to have done the former.

[next flax]

y

[vale wigeon]

you were supposed to divide the ENTIRE left-hand side by 4k

not cherrypick one term on it and divide only that by 4k and leave the rest to rot.

[next flax]

well there have been instances where the latter would be acceptable

where and how do i make the inference that the former is the right thing to do ?

[vale wigeon]

not literally as written there weren't.

when you divide both sides of an equation by something, you divide the ENTIRE lefthand side and the ENTIRE righthand side by that thing.

so in a sense the "former" in your symbolic conception thereof is always "the right thing to do"

even though the use of "the" to refer to something there need not always be exactly one of is questionable.

[next flax]

Here is an example of when I did the latter, and I got the correct answer

[vale wigeon]

you ADDED a thing to both sides

[next flax]

Ah shit lol

[vale wigeon]

addition isnt the same as division

youre trying to over-abstract

and you're falling flat on your face as a result

[next flax]

I think that’s reasonable

Lol

[next flax]

so wait (4k-10)/4k

is (2k-5)/2k ????

[vale wigeon]

sure, you can simplify (4k - 10)/(4k) into that

[next flax]

how would you simplify it

[vale wigeon]

for the purposes of solving this problem, it won't help you much.

but to answer your question as stated, you can simplify it all the way to $1 - \frac{5}{2k}$ maybe

[ocean seal]

Ann

[next flax]

i would've said,yeah

thats what i would've done

[vale wigeon]

(again, i must emphasize that this is losing relevance to the problem)

[next flax]

not really

i'm trying to understand how to get to -2.5

[vale wigeon]

well i can show you how to get from 4k - 10 = 8k to the solution while also doing your step of dividing both sides by 4k at the start

is that what you want

[next flax]

yeah! i would appreciate so much

[vale wigeon]

tbh i cant help but remark on the impression ive got from you
you seem to think algebra is procedurally rigid, like whenever you have an equation there's One and Only One Right Thing To Do,
and simplifications Must be done in One Particular Way and All Else Is Wrong,
and you, a mere mortal, are caught in the crossfire trying to understand wtf is going on

[next flax]

haha that's true

i just have preconceived notions on the way things work

[vale wigeon]

right

[next flax]

and i think i should appreciate prealgebra a bit more

[vale wigeon]

maybe it's good to unlearn those preconceived notions

[next flax]

because i think it would help me as you say

[vale wigeon]

algebra is kind of malleable

[next flax]

i have a goal of next summer i hope i can do abstract algebra

[vale wigeon]

like theres a big difference between a misstep and a fuckup

[next flax]

but that's a bit ambitious

[next flax]

whats the difference ?

[vale wigeon]

when you make an algebraic move that makes your working longer but is not by itself invalid, that's a misstep

when you make an invalid move that actually breaks something, that's a fuckup

fuckups happen only in execution

[next flax]

i see

because if i misstep

[vale wigeon]

a good step is just as liable as a misstep to get fucked up

[next flax]

i can correct it most likely

[vale wigeon]

yes, a misstep can be recovered from

[next flax]

unless i am on a cliff

then i'm probably fucked

well, a lot more risky

[vale wigeon]

in the worst of cases, you can backtrack to just before the misstep

i.e. erase your work and start anew

[next flax]

a misstep increases the probability of a fuck up, right?

[vale wigeon]

well

i guess insofar as your work becomes longer, yes

you have more places to fuck up in

also something to watch out for

if a step you're planning to take looks like it undoes the last step you took

it's probably a misstep

[next flax]

oh nvm

i see

[next flax]

how do they know k -3 = 9 without knowing the answer? :/

[abstract fractal]

They cross multiplied

If 6/(2k - 6) = 1/3, then 6(3) = 2k - 6

Dividing by 2, you get 3(3) = 9 = k - 3

[lone heart]

@next flax Has your question been resolved?

[vale wigeon]

they hid the step of dividing both sides by 2 after cross-multiplying

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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Available help channel!

Send your question here to claim the channel.

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

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

[tall current]

Can someone help me with 1 b please

[pine kettle]

what is the rate

after 40 minutes?

[tall current]

Yeah it’s asking what the rate is after 40 mins

[pine kettle]

what unit is V

[tall current]

m^3

[pine kettle]

ok

thats not necessarily a rate but a value

but we do know how to find the instantaneous rate of a function right?

(the derivative)

[tall current]

Yes it’s dQ/dT

[pine kettle]

okay

if we have V(t) (m^3)

then we can find V'(t) (m^3/minute) right?

[tall current]

Yes

[alpine sable]

whats $\frac{dV}{dt}$

[ocean seal]

Normal Cat

[tall current]

??

[alpine sable]

i mean to calculate the derivative

[alpine sable]

if u think then u will notice that its just V'(40)

[tall current]

hmm

[pine kettle]

we know V(t) so we can use our magic calculus powers to find V'(t)

[alpine sable]

whats a derivative

rate of change

[tall current]

Yeah I see it now

[alpine sable]

at a time

thats what we want

[tall current]

Yup

[alpine sable]

so calculate the derivative

[tall current]

okay I got an answer

Is it 6m^3 min^-1?

[alpine sable]

yup

[tall current]

ayyy

Thanks guys