help-33

[elfin berry]

pixel

[marsh citrus]

@low mica Has your question been resolved?

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[marsh citrus]

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[raw quest]

so for part a)

i got thiis

im not rly sure how to do part b tho

idk if this is the right idea but

i kinda grouped it so its like

(x^6+1/x^6) + 6(x^4+1/x^4)+(15(x^2+1/x^2)+20

im not sure wher to do from here tho

$(x^6+1/x^6) + 6(x^4+1/x^4)+(15(x^2+1/x^2)+20$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[low ruin]

seems correct

[raw quest]

oops extra bracket

but

then idk where do i go after this

[low ruin]

You just need equate the coefficients now

[raw quest]

um

im not rly sure how to get it into the form

[low ruin]

If 2x^2 -9x + 3 = Ax^2 + Bx + C

What are the values of A, B, and C

[raw quest]

$U^6 + AU^4 + BU^2 + C$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

um

then its

2, -9, and 3

[low ruin]

yep

[raw quest]

yep i got that

[raw quest]

$(x^6+1/x^6) + 6(x^4+1/x^4)+15(x^2+1/x^2)+20$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

but in this one

[low ruin]

Oh the form is slightly different I miss that

[raw quest]

so everything is in terms of U

right

where $$U = x+1/x$$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

im but u see here its like slightly different

[low ruin]

Why don't you move all the terms to the right of the equal sign? we need to make x^6 + 1/x^6 the subject

[raw quest]

yes

i tried that

so then i ended up with

$x^6 +1/x^6 = (x+1/x)^6 - 6(x^4+1/x^4) - 15 (x^2+1/x^2) - 20$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

and $U = x + 1/x$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[low ruin]

So we have U^6 - some_terms = x^6 + 1/x^6. So the form seems correct

[raw quest]

yes the only problem is

the powers r on the inside

its good for the first form

like

(x+1/x)^6

which is the same as U^6

[low ruin]

So we now have two problems to solve expressing x^4 +1/x^4 in terms of U and x^2 + 1/x^2 in terms of U

[raw quest]

but then im not sure what to do for the next ones

[raw quest]

yeahhh exactly

[low ruin]

Try to think about it. Maybe we could do something similar

[raw quest]

ive tried

thats why i came here to ask 😭😭😭

[low ruin]

!show

[marsh citrus]

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

[raw quest]

i couldnt figure out anything else

ive literally

gotten to this point

and now im lost

💀💀

[low ruin]

I will give you 10 minutes to think about it.

Problem solving is an important skill

[raw quest]

um

do u know how to do it

[low ruin]

I got idea but am not 100% sure

[raw quest]

oh ok

idk ive been

doing this for ike

30 mins

still not getting past this point

could u please share what u know

[low ruin]

Be creative. Maybe there is something you could do with U perhaps expansion

[raw quest]

idk

i rly dont know

<@&286206848099549185> pls help 🙏

[balmy magnet]

what is the problem type

[raw quest]

binomial expansion

wait just part b tho

[low ruin]

Expand (x+1/x)^2

[balmy magnet]

I see

[low ruin]

Also consider (x+1/x)^4

[raw quest]

part a was fine

[balmy magnet]

yea a is ez

[raw quest]

so

should i try it

and then

like

minus the excess?

[low ruin]

yep

[raw quest]

ohhh

okok lemme give that a go

[balmy magnet]

replace U with the values and solve the equation as right-handed side = 0

ig

[raw quest]

bruh

i dont think its orking

wait

nvm

i got it

holy

that toook a while

@low ruin thanks for suggesting the removing excess

got it now 🙏

.close

[marsh citrus]

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[marsh citrus]

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[honest current]

Where have they gotten 9x2^2k

[turbid nacelle]

since k>=9

then k+1>=10

and as such if you say that k+1>9 it is reasonable

[marsh citrus]

@honest current Has your question been resolved?

[honest current]

I still don’t understand

[tacit fractal]

there is a number which is greater than 10

tell me such a number

[honest current]

11

[tacit fractal]

okay

Can't this number be greater than 9?

[honest current]

Yes it can

[tacit fractal]

so yes if a number is greater than 10, it is also greater than 9

if u can write k > 10

u can also write k > 9

[honest current]

But it doesn’t mention greater than 10

[tacit fractal]

k+1 > 10

k + 1 = apple

apple > 10

apple > 9

(k+1)×2^2k > 9 × 2^2k

(k+1) > 9

2^2k's goes

[tacit fractal]

.

apple > 9

because it said that apple > 10

i said apple because it is just a number

i just used this as a variable

k+1 is a number after all

which is greater than 10

so k+1 also greater than 9

so you can write this: (k+1)×2^2k > 9×2^2k

because it says k+1 > 9

After 2^2k's are simplified

there is just k+1 > 9

which is not a wrong impression

because we knew that k+1 > 10

[honest current]

Ugh Idk why it’s not

Clicking

In my head

The whole thing has confused me

Like the steps they’ve done to prove it

[tacit fractal]

what is explained here

proof of what

[honest current]

Prove the statement

I mean the question

Like

I don’t understand why they’ve done what they’ve done

[tacit fractal]

n! > 2^2n?

n > any number

[tacit fractal]

with that condition

ig they are trying for n ≥ 9

n ≥ 9,

prove that n! > 2^2n

is that all?

[low ruin]

When going to some question requiring to prove something. Often we don't know what we are doing as result we just need to give it a go

[honest current]

I don’t understand why they’ve multiplied 9 by 2^2k

Why have they done it and how

Like the purpose

[low ruin]

It just one part of steps as mention above ("why"). However we can examine how

So we have $(k+1)k! > (k+1)2^{2k} > 92^{2k}$

[elfin berry]

team132

[low ruin]

from the assumption we get the first inequality

For the second inequality we assume the statement holds k = 9

If k = 9 then k+1>9

[tacit fractal]

we find again what we assumed, it seems strange to me

like if you assume k! > 2^2k, k ≥ 9

(k+1)! ≥ 2^(2(k+1)) is expected to be

i don't know it seems strange to me

maybe i am wrong

[low ruin]

For induction we assume statement is true for k (this comes from base case). We then prove that if the statement is true for n = k then it is true for n = k+1

[tacit fractal]

What am i trying to prove?

here

[low ruin]

k! > 2^2k

[tacit fractal]

oh i see

it was written above

ok we assumed this is true, then what?

[low ruin]

We then prove that if the statement is true for n = k then it is true for n = k+1

[tacit fractal]

why do we need something ridiculous like n

we have k already which looks so cool

like n = k

like dude i did not change such a variable even when doing integrals

XD

this sounds ridiculous

n = k

what is n?

[low ruin]

n is a variable

[tacit fractal]

yes k is a variable too, why are we bothering with two of them

can't we use just k

if we prove that it is true for k, we already prove that it is true for n

because n = k

[low ruin]

Initially our statement use n. Since dealing with k and k+1 we need two variables.

[low ruin]

The symbol '=' is not really an equal sign in this context. It is more of an assignment operator

n represents a particular value for this which the theorem holds

[tacit fractal]

okay let's continue

i still don't understand what to do with these two situations

how do we know that it provides for k+2 after seeing that it provides for k+1

because we need all integers greater than 9

Although after "k! > 2^2k, k ≥ 9" proving this the others are also proven

but i don't understand our purpose

[tacit fractal]

i expected to understand something when i read this

but i couldn't

@low ruin can you continue the proof? if you are available

maybe i can understand from you

[marsh citrus]

@honest current Has your question been resolved?

[marsh citrus]

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[sharp mauve]

how did the tan appear?

[open plume]

$\tan(\psi) = \frac{\sin(\psi)}{\cos(\psi)}$

[elfin berry]

lpieleanu

[sharp mauve]

.close

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[charred panther]

Asked the same question yesterday, but not sure if the approach that was suggested then is the one I'm "supposed" to take in my math class

[charred panther]

Find the orthonormal basis to the null space/room for the matrix A

So what I did was the following

x+y+z+w = 0
x+2y+2z = 0
y+z-w = 0

solving that i get

x+y+z+w = 0
  y+z-w = 0
 0*z+0*z= 0
which gives

x = -2s
y = s-t
z = t
w = s

If my understanding is correct, s and t gives the bases for the nullspace

so we have s(-2, 1, 0, 1) and t(0, -1, 1, 0)

and this is the part where I am so confused

The normating part is not my issue, there we get sqrt of -2^2+1^2+1^2

aka sqrt(6) for s

and t is sqrt(2)

but s scalar t isn't 0, which means they're not orthogonal

and the proposed solution (gram-schmidt) isn't in my course literature

and in fact, I am not able to find anything that would help me solve this..

my understanding is that A is a matrix with 4 vectors from r^3

But what does it then mean if we get a null space with 2 vectors with 4 numbers in them?

And if so, how could I make them orthogonal?

Checking the "solutions" (such a joke lmao) they get the same basis for the null space/room as I do, but they then get that an orthonormal base is 1/sqrt(2) (0,-1,1,0) and 1/sqrt(22) (-4, 1, 1, 2)..

What I do see is that you can get the second basis by doubling the s vector and then adding t to that, but I have no idea why that is happening :(

aka

first base is s(-2, 1, 0, 1)
second base is t(0, -1, 1, 0)

extending first base with 2
2*s(-2,1,0,1) = (-4, 2, 0, 2)

adding first base to that gives
(-4,2,0,2) + (0,-1,1,0) = (-4, 1, 1, 2)

[marsh citrus]

@charred panther Has your question been resolved?

[red cosmos]

true

[charred panther]

<@&286206848099549185>

Think I was allowed to ping after 15 minutes

[marsh citrus]

@charred panther Has your question been resolved?

[charred panther]

.c.ose

.close

[marsh citrus]

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[pliant reef]

not on a specific question but how to in general solve questions like
given that length AC is [some number] OR given that points ABC form a triangle with [some number] area, determine the 2 possible locations of point C

[cedar smelt]

if you know AB and area

AB*hC/2 = area

so hC= 2Area/AB

[pliant reef]

sorry I mistyped my question its when AC is equal to some number find 2 possible locations of point C

.close

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[rose mica]

for all n>1 ?

then yea, its a constant sequence

[stark bramble]

Yes

[copper raven]

what's it though ?

a_n itself

or something else

[marsh citrus]

@still temple Has your question been resolved?

[uncut oracle]

wht to do after first diff

[copper raven]

@uncut oracle i'll go in your channel

[marsh citrus]

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[still temple]

why is this not continuous?

[twilit grove]

not defined at x=1

[still temple]

but u can factorise top and cancel

[twilit grove]

it's still not defined at x=1

[still temple]

so for continuity u just sub in number

and if it doesnt exist its not cts?

the graph looks cts to me

surely the question is wrong and the graph is actually continuous

[twilit grove]

what is f(1)?

[still temple]

2

[twilit grove]

no

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

what is f(1)

[still temple]

so the answer is yes then

u do just sub it in

limits dont matter

f(x) = (x^2 - 1) / (x - 1)
= ((x + 1) (x - 1)) / (x - 1)
= x + 1

how is that not correct

[still temple]

this shi dont make no sense to me man

[twilit grove]

the limit as x approaches 1 equals 2, yes

but f is not defined at x=1

so it is not continuous

[still temple]

why cant i just write f(x) as x + 1

if it simplifies to that

[twilit grove]

it is still not defined at x=1

there is a "hole" there

[still temple]

x + 1 is defined at 1 tho, if the function can be simplified to x + 1 why is it wrong to write that

[twilit grove]

A function $f$ is continuous at a point $a$ if $\lim_{x \to a} f(x) = f(a)$

[elfin berry]

cwatson

[twilit grove]

for $a = 1$, $f(a)$ is not defined. it doesn't matter that $\lim_{x \to 1} f(x) = 2$. there is no $f(1)$.

[elfin berry]

cwatson

[still temple]

ok ty

.close

[marsh citrus]

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[kindred hinge]

I need help with limits at infinity

[broken dome]

What problem are you dealing with exactly?

[kindred hinge]

I need help with working with indeterminate forms

[broken dome]

Ok, could you provide an example of a problem you've had issue with?

[kindred hinge]

like I know I can use l'hopitals for any 0/0 or inf/inf, but what about when you plug something in and you have inf - inf? the example being lim x -> inf (10x+14+6x^2-x^4) = ?

[kindred hinge]

^

[dapper badge]

like an integral, you can find limits by term

[kindred hinge]

how so?

[dapper badge]

such as

[vital oracle]

trying to find the limit by term of 10x + 14 + 6x^2 - x^4 is what gets you ∞ - ∞ to begin with, so that wont always work
you can look to cancel things out, but you edited the question so that that doesnt happen

theres a rough tactic you can use in 0/0, ∞/∞, ∞ - ∞ where you "choose the bigger infinity"

[kindred hinge]

How does that work?

[vital oracle]

have you heard of big O notation before

[kindred hinge]

ive heard of it, but ive never formally learned it

[vital oracle]

you consider O(10x + 14 + 6x^2 - x^4)

that goes down to O(x^4)

therefore you consider the -x^4 term as the 'largest', and do lim -x^4 instead

this rough tactic is the easiest but the roughest

[kindred hinge]

got it

[vital oracle]

if you want a less rough version,

begin with the ∞ - ∞ youre stuck with

consider ln(e^(∞ - ∞))

then thats ln(e^∞/e^∞)

in your case thats ln(e^(10x+14+6x^2)/e^(x^4))

youll notice that inside the ln is essentially a ∞/∞ that you can use l'hopital on

this is noticibly harder but that way you can convert a ∞ - ∞ into a ∞/∞

[kindred hinge]

That is really clever

let me think about it for a second

wouldnt you end up with just getting e^-infinity/ e^infinity?

[vital oracle]

the top is e^∞ not e^-∞

[kindred hinge]

I thought the derivative of ln was the reciprocal of its argument

[vital oracle]

oh youre doing l'hopital

also

thats not l'hopital

shouldnt l'hopital require a fraction?

[kindred hinge]

it does

[vital oracle]

and whats inside the ln

[kindred hinge]

got it

to be honest im still a little lost

[vital oracle]

what does your limit look like so far

[kindred hinge]

it is looking like zero over zero, which I think is wrong

[vital oracle]

do you know what 0/0 means

[kindred hinge]

the limit doesnt exist

[vital oracle]

no

[kindred hinge]

what does it mean?

[vital oracle]

0/0 is an indeterminate form

[kindred hinge]

right

[vital oracle]

an indeterminate form means that youre not done with the limit yet, you havent determined a number

also shouldnt you be getting ∞/∞?

[kindred hinge]

let me go take a photo real quick

[vital oracle]

btw when you finish getting this limit, Ill show a polynomial-specific way to find this limit

[kindred hinge]

thank you

[vital oracle]

bruh

[kindred hinge]

I got -inf/inf, I dont know why I said zero sorry

[vital oracle]

listen

[kindred hinge]

yes?

[vital oracle]

look at this

what do you think I meant when I typed that

[kindred hinge]

Im a dumbass

let me retry it rq

[vital oracle]

theres so many things wrong with the work youve done

you look like youre going to make the same mistakes again

[kindred hinge]

idk why I subbed it in so early

[kindred hinge]

sorry

[vital oracle]

thats not the reason

see youre not thinking about why 0/0 and ∞/∞ are called indeterminate forms to begin with

where do you think youre heading from this

"subbing it in early" just sounds like "I made a mistake in a method I dont understand" instead of "Ive just done something that feels very wrong"

[kindred hinge]

its both on my end, where can I find some good resources so I can adequately understand the method?

[vital oracle]

Im right here you know

how about I reteach you this?

[kindred hinge]

that would help

[vital oracle]

it would only help?

[kindred hinge]

it would be a life saver actually

[vital oracle]

you need to put more thought into your words

we're going to work with ∞ here so words are important to get right

earlier I converted ∞ - ∞ into ln(e^∞/e^∞)

thats a shorthand, a figure of speech

that stands for "convert this - that into ln(e^this / e^that)"

or $\lim_{x\to\infty}(10x+14+6x^2-x^4)=\ln\qty(\lim_{x\to\infty}\frac{e^{10x+14+6x^2}}{e^{x^4}})$

[elfin berry]

mtt

[kindred hinge]

that seems obvious now, I dont know why I didnt catch that

[vital oracle]

its only obvious now because you thought ∞/∞ was going to help

you didnt take the fundamental problem with ∞/∞ seriously

[kindred hinge]

because we're doing this to get it out of that form in the first place

[vital oracle]

we're not discussing method right now

[kindred hinge]

if its still in that form, its obviously wrong

[vital oracle]

you know the reason?

[kindred hinge]

please give me a second to think

[vital oracle]

if you dont get this, I can reteach you the reason

try your best to be specific

[kindred hinge]

since an indeterminate form means that we havent determined a number for the limit, it (infinity/infinity) doesnt adequately describe what the function is approaching

[vital oracle]

thats pretty good

and what is it about dividing ∞ by ∞ that turns out vague?

if you do ∞/1, that = ∞
if you do 0/∞, that = 0

you can view this as shorthand for lim (a function that tends to ∞) / (a function that tends to 1) = ∞

[kindred hinge]

its that there is no way inf/inf can be equal to one, since its infinitely dividing itself

[vital oracle]

thats not correct

[kindred hinge]

right

[vital oracle]

,,1=\lim_{x\to\infty}\frac xx=\frac{\infty}{\infty}

[elfin berry]

mtt

[vital oracle]

,,2=\lim_{x\to\infty}\frac{2x}x=\frac{\infty}{\infty}

[elfin berry]

mtt

[vital oracle]

you were correct that an indeterminate form does not have enough to determine a number

here, you can see that ∞/∞ cannot equal just 1 or just 2, you need more information on where those ∞s come from

the fraction that you got those ∞/∞s from would be that information

thats what youre doing a limit for

[kindred hinge]

right

[vital oracle]

however we know that ∞/∞ doesnt work but 0/∞ does

consider (function that tends to ∞) / (a different function that tends to ∞)

if the numerator "grows faster" than the denominator, it will dominate and you get ∞

if the denominator "grows faster" than the numerator, it will dominate and you get 0

between those, if the denominator and numerator grow equally, you get a finite number

the idea here is, like ∞ - ∞, both infinities are working against each other

and so which one wins requires more context

usually people think ∞ / ∞ = 1 because they'd think the same function is being placed for both ∞s since the same symbol ∞ is being used

[kindred hinge]

let me think about that for a second

[vital oracle]

ok

[kindred hinge]

what do you mean by growing faster? like the slope?

[vital oracle]

youll need to know more about big O notation

we now need to teach power

think of $\lim_{x\to\infty}\frac{x^2}x$

[elfin berry]

mtt

[vital oracle]

you would expect this to = ∞, right

[kindred hinge]

right, since our numerator grows faster than our denominator

[vital oracle]

bruh

[kindred hinge]

what did I miss?

[vital oracle]

look closer at this image and tell me if theres something off with you saying "grows faster"

growing faster has nothing to do with slope btw

[kindred hinge]

it should be equal to zero since we have a vertical asymptote at x = 0

[vital oracle]

🤦‍♂️

the problem here is youre asking about "grows faster" as if you want a better sense of how to use that word

you then proceed to use "grows faster" regardless

you used it the very moment I mentioned an example btw

so you inadvertently lied here or there

[kindred hinge]

lied how?

[vital oracle]

I didnt say you lied

I said you INADVERTENTLY lied

that means you didnt communicate what you were saying properly

you accidentally didnt say what you meant

surely I dont need to explain "inadvertently" to you man

screw this

[kindred hinge]

okay, thank you for trying

[vital oracle]

we're skipping straight to that polynomial thing I mentioned earlier

wait

you wouldnt know how that works

because you dont know ∞ * 0

you see the issue here?

[kindred hinge]

im missing alot of pre-req knowledge

[vital oracle]

yep so we're in fact sticking to this until you get this

as soon as that's built up, then we move on

[kindred hinge]

got it

[vital oracle]

do you get this

oh btw Ive come up with a different way of saying what indeterminate forms are, it should work

[kindred hinge]

I dont think I actually do

[vital oracle]

yep so lets stick to some examples

but before we do that, we need to address two things

[vital oracle]

first

the function is x^2 / x

and the limit is x -> ∞

so what do you mean by "it should be 0 because x^2 / x has a vertical asymptote at x = 0"

[kindred hinge]

I dont think I thought that through as much as I should have, since the limit is dependent on the y-values not the x's

it should be infinity

[vital oracle]

does x^2 / x have a vertical asymptote?

[kindred hinge]

it does, since 1/x at zero must have a vertical asymptote, because 1/0 is undefined

the value of x^2 doesnt affect where the vertical asymptote is

[vital oracle]

you dont think that the x^2,

which is x * x,

or denominator * denominator,

can affect anything in denominator * denominator / denominator?

[kindred hinge]

it should be just the denominator

[vital oracle]

x^2 / x = x

here's a graph of y = x^2 / x

theres no asymptote

what you said about 1/x -> ∞ as x -> 0 is a rule for when theres a constant in the numerator

[kindred hinge]

right, I think I was overthinking it and confusing it for when theres (x^2+c)/x

[vital oracle]

you can view a discontinuity as the case of "0/x when x -> 0"

its a "vertical asymptote with no width"

none of that's really true you-wise

for now you can consider the point here as a discontinuity that can be simplified

[vital oracle]

now here's the second thing

[vital oracle]

if youre still stuck on this, we can explain it afterwards, it means something weird

for now though you need the basics

[kindred hinge]

what I know about vertical asymptotes, its where all y's are undefined

[vital oracle]

thats not what vertical asymptotes are often called

[kindred hinge]

where x is in the denominator and equal to zero

[vital oracle]

I cant say yes or no to that without getting into the weird mentioned here

[vital oracle]

lets stick to this for now

ok?

[kindred hinge]

ok

[vital oracle]

first, you seem to have forgotten that keyboards often type different kinds of math because they only get to use keyboard symbols

do you think the ^ should be on your paper

[kindred hinge]

No, I have no clue why I did that

[vital oracle]

you did it four times

in a row

youll need to address what caused that

(on your own)

theres also something else,

whats happening here

I cant make out what you wrote down

[kindred hinge]

something stupid, its when I took your shorthand literally

[vital oracle]

to be more specific, you wrote an x in the denominator

is this $\ln\qty(\frac{e^{\infty}}{e^{\infty}})$

[elfin berry]

mtt

[kindred hinge]

that is what I meant

but the denom was e^ negative infinity

[vital oracle]

no it wasnt

exponent laws say that $e^{a-b}=\frac{e^a}{e^b}$

[elfin berry]

mtt

[vital oracle]

you can see there the b loses the - sign

I showed you this and I hoped you wouldve noticed (twice)

[kindred hinge]

oh so its like the quotient property of logs

[vital oracle]

bit concerning that you went with that

surely theres an easier way to know this law

[vital oracle]

is this your first time seeing this law

[kindred hinge]

No, we learned them in algebra two

[vital oracle]

youre correct that its very much like the log quotient property, but its not a good basis if you want to remember the law as making sense

usually its (sense) -> (exponent law) -> (log law)

the idea is if you have e^5 / e^3,

then thats (e * e * e * e * e) / (e * e * e), of which e * e is remaining

[kindred hinge]

right

[vital oracle]

so in general e^(a - b) = e^a / e^b

[kindred hinge]

which is the same as e^ 5-3

[vital oracle]

similar to how e^(a + b) = e^a * e^b

[kindred hinge]

got it

[vital oracle]

while we're here, you can plug in a = 0 and b = 1 to get that
e^(0 - 1) = e^0 / e^1
or
e^-1 = 1 / e

so that would show that x^-1 = 1/x

[kindred hinge]

right

[vital oracle]

not really relevant but its neat that it can come from e^(a - b) that quickly

alr brb
go explain what this means in the meantime

[kindred hinge]

its bad notation for derivative with respect to x

it should be d/dx

[vital oracle]

yea and it should be before, not after

[kindred hinge]

you only do it after for integration right?

[vital oracle]

sure

alr I gtg for now but Ill be back in maybe 30min

we've gone over all the errors in the picture now

[kindred hinge]

thank you, ill rework it while you're out

[vital oracle]

alr

oh btw

for the polynomial-specific method, have the first step be forcibly factoring an x^4 out of the expression

[kindred hinge]

alright

[marsh citrus]

@kindred hinge Has your question been resolved?

[kindred hinge]

ok so while trying to do it this way, I came across a problem, since e^(polynomial) derives into e^(polynomial) * (d/dx polynomial), how does it reduce to a determinate form?

[vital oracle]

you didnt write down this d/dx correctly

[kindred hinge]

it should be two seperate ones, my bad

[vital oracle]

other than that yea I dont think this is getting anywhere

so the polynomial way would be the only way you oculd write down

try that instead:

,,\lim_{x\to\infty}(10x+14+6x^2-x^4)=\lim_{x\to\infty}(x^4)\qty(\frac{10}{x^3}+\frac{14}{x^4}+\frac6{x^2}-1)

[kindred hinge]

ill start on it

[elfin berry]

mtt

[vital oracle]

ok maybe we need to continue with the basics

[kindred hinge]

wait, any constant divided by infinity is zero right?

I see where I messed up

[vital oracle]

yep thats correct

[kindred hinge]

awesome

thank you for putting up with me, im especially sorry for putting carets on paper lol

[vital oracle]

lol np

did you want to relearn the basics

[kindred hinge]

it couldnt hurt, I really want to be sure I actually understand this

[vital oracle]

nice alr

in calculus, ∞ is a shorthand for "a function that approaches infinity"

[kindred hinge]

right

[vital oracle]

so lets say you have ∞/∞, this represents dividing a numerator and denominator that are both approaching ∞

call these functions $\lim_{x\to a}\frac{n(x)}{d(x)}$

[elfin berry]

mtt

[kindred hinge]

right

[vital oracle]

now you know from earlier that 0/∞ = 0, this represents that lim (function -> 0)/(function -> ∞) is always a limit that equals 0

[kindred hinge]

right, since zero is a constant, and any constant divided by infinity must be zero

[vital oracle]

thats not what I said

instead 0/∞ = 0 represents that lim (function -> 0)/(function -> ∞) is always a limit that equals 0

[kindred hinge]

let me make sure im interpreting this correctly, so the limit as the x of n(x) approaches zero divided by the limit as x of d(x) approaches infinity must always equal zero

or is there something im not getting

[vital oracle]

if $\lim_{x\to a}n(x)=0$ and $\lim_{x\to a}d(x)=\infty$, then $\lim_{x\to a}\frac{n(x)}{d(x)}=0$

[elfin berry]

mtt

[vital oracle]

you got the idea

[vital oracle]

now you can see there that "0/∞" is pretty well defined

[kindred hinge]

I see

[vital oracle]

maybe to decrease confusion, any time I do "0" in quotes, its short for "a function that approaches 0" or "a function whose limit is 0"

so lim "0"/"∞" = 0

[kindred hinge]

got it, so it will always refer to the limit of the corresponding function

[vital oracle]

yep

same with "∞" also

that way we're only ever see the bare ∞ in rare situations

[kindred hinge]

got it

[vital oracle]

now we expect lim "0"/"∞" = 0

lets see if lim "∞"/"∞" has a specific value

we can check what happens for various examples of n(x) and d(x)

for now, we're doing that $\lim_{x\to a}n(x)=\infty$ and $\lim_{x\to a}d(x)=\infty$

[elfin berry]

mtt

[vital oracle]

so an example of this is when n(x) = 2x and d(x) = 2x (and a = ∞)

now $\lim_{x\to\infty}\frac{2x}{2x}=\lim_{x\to\infty}1=1$

[elfin berry]

mtt

[kindred hinge]

that makes sense, since the terms share a common variable, so all its reduced to is a constant

[vital oracle]

now consider $\lim_{x\to\infty}\frac{cx}{x}=\lim_{x\to\infty}c=c$

[elfin berry]

mtt

[vital oracle]

this is an example of n(x) -> ∞, d(x) -> ∞ where you can get almost any real number c from "∞"/"∞"

[kindred hinge]

this looks identical to the 2x/2x scenario but instead its just 2x/x, leaving just 2 or in this case c

[kindred hinge]

never mind, its because the x cancels always leaving c

[vital oracle]

yea

[vital oracle]

now through this, you can see that "∞"/"∞" can equal any positive number

[kindred hinge]

why can't c be negative?

[vital oracle]

because we had to assume to begin with that n(x) -> ∞ instead of n(x) -> -∞

[kindred hinge]

so c must have a consistent signage with the limit the corresponding function is approaching?

[vital oracle]

not really sure if I should confirm that

[vital oracle]

the limit here shows a corerct calculation for this particular fraction

which indeterminate form the calculation is depends on the value of c

if c is positive, cx/x equals the indeterminate form ∞/∞

if c is 0, cx/x equals 0/∞ which is not indeterminate

if c is negative, cx/x equals the indeterminate form -∞/∞

however ∞/∞ and -∞/∞ are often considered the same indeterminate form

similar to how 2 * ∞/∞ and ∞/∞ are still considered as a general "infinity over infinity"

[kindred hinge]

right

[vital oracle]

youre looking for the general form it takes on

[kindred hinge]

can you give me just a second to write this down^

[vital oracle]

ok

[kindred hinge]

I got it down

[vital oracle]

alr

now notice that "∞"/"∞" can equal any positive number

because we created an example of that for lim cx/x

[kindred hinge]

right

[vital oracle]

now "∞"/"∞" can also equal ∞ or 0, it depends on choosing differing kinds of n(x) and d(x)

examples are:

,,\lim_{x\to\infty}x^2=\infty,\lim_{x\to\infty}x=\infty,\lim_{x\to\infty}\frac{x^2}{x}=\lim_{x\to\infty}x=\infty
\\lim_{x\to\infty}x=\infty,\lim_{x\to\infty}x^2=\infty,\lim_{x\to\infty}\frac{x}{x^2}=\lim_{x\to\infty}\frac1x=0

[elfin berry]

mtt

[kindred hinge]

I understand, so as long as there exists at least one x without a denominator (assuming no different signed x's), the limit as x approaches infinity must be infinity, and that for any constant over x as it approaches infinity must be zero

[vital oracle]

sure but dont form that pattern too strongly

n(x) and d(x) are more general functions

the point is what "∞"/"∞" can evaluate to

instead of "how to simplify x^2/x"

[kindred hinge]

got it

and just to reiterate, "inf" means the limit as x of the corresponding function approaches inf

[vital oracle]

yea

so "∞"/"∞" can be 0, a positive number, or ∞

[kindred hinge]

got it

[vital oracle]

for this raeson, ∞/∞ is called an indeterminate form

[kindred hinge]

since it can be any of those 3

[vital oracle]

yea

this can mean "if you get a limit that evaluates to ∞/∞, then youre not done figuring out the limit"

this can also mean "by itself, ∞/∞ doesnt tell you enough to equal a number, so its indeterminate"

[kindred hinge]

that makes alot of sense now

[vital oracle]

now observe that ∞/∞ is happening depending on how "strongly" n(x) and d(x) approach ∞

[kindred hinge]

how would you define strongly?

[vital oracle]

usually its dividing n(x)/d(x) to verify how comparatively strong n(x) and d(x) are

so you can see its a bit circular to get the sense of "strongly"

the definition is if $\lim_{x\to\infty}\frac{n(x)}{d(x)}=0$, then $d(x)$ is stronger or grows larger or grows faster than $n(x)$

[elfin berry]

mtt

[kindred hinge]

does this imply that d(x) likely has a greater exponent on x compared to n(x)?

[vital oracle]

you can consider it that way

formally this is called "d(x) dominates n(x) asymptotically"

so all of our sense of bigger/stronger/larger/faster is better described as "asymptotically"

[kindred hinge]

what do you mean by asymptotically? im assuming it just meants that d(x) is in the denominator, but im not sure if thats the case

[vital oracle]

I said definition

that means I just told you what asymptotically meant

[vital oracle]

put it in quotes even

this is just what its called

formally

[kindred hinge]

^

my bad

I dont know how I missed that

[vital oracle]

brb

[kindred hinge]

alright

so just to make sure I understand, if a function dominates another function asymptotically, it must mean that the first function has greater strength compared to the other, which we can consider the first function to have a leading term whose exponent is greater than the second function's leading term's exponent

[marsh citrus]

@kindred hinge Has your question been resolved?

[vital oracle]

back

you here?

[kindred hinge]

yes

[vital oracle]

great youre back

[vital oracle]

polynomials arent the only functions out there

if n(x) and d(x) are both polynomials, youd be right

[kindred hinge]

how would you define strength for some of the non-polynomials?

[vital oracle]

let me repeat

if $\lim_{x\to\infty}\frac{n(x)}{d(x)}=0$, we say $d(x)$ "dominates $n(x)$ asymptotically"

[elfin berry]

mtt

[vital oracle]

for example $\lim_{x\to\infty}\frac{x}{e^x}=0$, so $e^x$ dominates $x$ asymptotically

[elfin berry]

mtt

[vital oracle]

"d(x) dominates n(x) asymptotically" is equivalent to saying that
"d(x) grows faster than n(x)"
"d(x) grows larger than n(x)"
"d(x) is stronger than n(x)"

[kindred hinge]

I have a question

so if d(x) grows faster than n(x), doesnt that mean that the derivative of d(x) must be larger than n(x)'s?

[vital oracle]

to be more specific, are you asking that:

,,\lim_{x\to\infty}\frac{n(x)}{d(x)}=0\implies\lim_{x\to\infty}\frac{n'(x)}{d'(x)}=0?

[elfin berry]

mtt

[kindred hinge]

let me think for a second

[vital oracle]

also again youre thinking of slope

that is incorrect

this has nothing to do with derivatives

asymptotic dominance has nothing to do with slope

[kindred hinge]

so what do you mean by growth here?

[vital oracle]

I already told you

[vital oracle]

it specifically means the outcome of this calculation

THIS is your definition

this is the third time youve seen it

do you need me to tell you it a fourth?

[kindred hinge]

no

[vital oracle]

its going to be a yes if youre not careful

are you sure?

[kindred hinge]

I keep making a false equivalency, im sorry

[vital oracle]

lets address derivatives then

growth has two separate definitions

the first one is the derivative

[vital oracle]

the second one is this

[kindred hinge]

got it

[vital oracle]

now we're talking on a more fundamental scale than just derivatives

you already have a sense that x^100 is in a sense stronger or fundamentally steeper than x^2

[kindred hinge]

right

[vital oracle]

its not the case of what the slopes are

its a case of how the function behaves

and that x^100's shape is fundamentally different and will always surpass x^2

also like how x surpasses 1/x

[kindred hinge]

I see

[vital oracle]

you could multiply 1/x by a constant

you could have it be 999999/x

but that doesnt change that its shape fundamentally goes down compared to x

that it has to shrink compared to x

[kindred hinge]

im really trying to wrap my head around this

[vital oracle]

youre not looking for values

youre comparing shapes

youre considering that one of them is an x while the other is merely a 1/x with a constant

this is where o notation starts to appear, also

asymptotic dominance is like a <

so 1 = o(x)

[kindred hinge]

what is n subscript o?

[vital oracle]

maybe you shouldnt be reading what you dont understand

if we didnt address it yet

thats a 0

not an o

do you want the formal definition now? I still havent told you how to do f(n) = o(g(n)) yet

[kindred hinge]

not yet

[vital oracle]

when you get that lim x -> ∞ f(x)/g(x) = 0,

you say that f(x) = o(g(x))

for example, 1 = o(x) because lim x -> ∞ 1/x = 0

and x = o(e^x) because lim x -> ∞ x/e^x = 0

[kindred hinge]

I see

[vital oracle]

you can see that using an = sign here is a bit off

o(g(x)) is only as precise as saying "a function that is asymptotically dominated by g(x)"

you can see it isnt very precise

for example, all you would know about a function f(x) = o(x^2) is that lim x -> ∞ f(x)/x^2 = 0

that can still allow a lot of functions, like f(x) = x^1.99 or f(x) = 1

[kindred hinge]

right, since f(x) is unknown here

[vital oracle]

its like saying "even + odd = odd" for numbers

[kindred hinge]

its not exact but it explains what its trying to explain

[vital oracle]

alr if you cant notice asymptotic dominance intuitively, we'll just leave it as this particular fact

you can see that using this acts as a < sign

theres a very similar version that acts as a > sign

[kindred hinge]

alright

[vital oracle]

$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\pm\infty$ means that "$f(x)$ dominates $g(x)$ asymptotically"

[elfin berry]

mtt

[vital oracle]

youll notice that all of the statements above can be flipped to use this version instead

$\lim_{x\to\infty}\frac{e^x}{x}=\infty$, so $e^x$ dominates $x$ asymptotically

[elfin berry]

mtt

[vital oracle]

$\lim_{x\to\infty}\frac{x}{1/x}=\infty$, so $x$ dominates $\frac1x$ asymptotically

[elfin berry]

mtt

[vital oracle]

$\lim_{x\to\infty}\frac{x}{1}=\infty$, so $x$ dominates $1$ asymptotically

[elfin berry]

mtt

[vital oracle]

heres an equivalent symbol for that

f(n) = omega(g(n))

so x = omega(1), e^x = omega(x)

1 = o(x) and x = omega(1) mean the same thing

like 1 < 3 and 3 > 1

or more specifically 1 = (number less than 3) and 3 = (number greater than 1)

[vivid breach]

too much math for me

[kindred hinge]

let me think about this for a second

[vital oracle]

dont read into it too closely

[kindred hinge]

ok

I think im about done for today

[vital oracle]

alr

you didnt get enough of this

theres a list of which functions dominate which others

do you want to see that

[kindred hinge]

sure

[vital oracle]

youll need to know this in calculus

lim x -> ∞ stronger ones / weaker ones = ±∞

[kindred hinge]

its better sooner than later

[vital oracle]

lim x -> ∞ weaker ones / stronger ones = 0

one more thing, if you get lim x -> ∞ n(x)/d(x) = (a nonzero finite number), n(x) and d(x) are "equal asymptotically"

here the image wants the limit to = 1, but its not that big of a deal

the idea here is that x ~ 2x because theyre both lines

you expect them to behave similarly and be "equally strong"

so now we have a <, an =, and a >

which is o, ~, and omega

alr heres the list

first theres constants, like f(x) = 1, f(x) = 2, or f(x) = -3

[kindred hinge]

right

[vital oracle]

above that theres f(x) = x^(a power)

in a polynomial, you consider the highest power

same with x^1.5 + x^0.5 for example, you consider x^1.5

higher powers > lower powers

[kindred hinge]

got it

[vital oracle]

so √x, x, x^2, x^3, x^4, etc.

in that order

[kindred hinge]

so basically, the higher the degree, the higher the order, and the higher the strength

[vital oracle]

sure

I think "order" is the best normal word you can pick

higher and lower orders

this is just a more detailed way to use degrees on functions that arent just polynomials

now heres an interesting one

ln(x) is between constants and x^(a power)

so $\lim_{x\to\infty}\frac{\ln(x)}{x^{0.001}}=0$

[elfin berry]

mtt

[kindred hinge]

??????????

[vital oracle]

do you want a proof?

[kindred hinge]

sure

[vital oracle]

before we do that, we need to consider a more believable example

exponentials like e^x are more powerful than any polynomial

so $\lim_{x\to\infty}\frac{e^x}{x^{100}}=\infty$

[elfin berry]

mtt

[vital oracle]

this limit up here, you can prove if you consider doing l'hopital 100 times

the top is still e^x

the bottom becomes a constant because x^100 -> 100x^99 -> ... -> a constant

[kindred hinge]

so is strength determined by the number of derivatives you can use on it??

[vital oracle]

no

thats still called degree

again I told you

[kindred hinge]

Right

[vital oracle]

strength is determined by $\lim_{x\to\infty}\frac{f(x)}{g(x)}$

[elfin berry]

mtt

[vital oracle]

if this is 0, its like a <

if this is ±∞, its like a >

if this is any number in between, its like an =

[kindred hinge]

right, so what exactly makes e^x more powerful than any polynomial?

[vital oracle]

exponentials like e^x are more powerful than any polynomial

so $\lim_{x\to\infty}\frac{e^x}{x^{100}}=\infty$

[elfin berry]

mtt

[vital oracle]

this limit up here, you can prove if you consider doing l'hopital 100 times

the top is still e^x

the bottom becomes a constant
each time you d/dx, you decrease the power by 1
so d/dxing 100 times would decrease the power from 100 all the way to 0
the 100th derivative of x^100 is just 100 * 99 * 98 * ... * 3 * 2 * 1, a constant

[kindred hinge]

I understand that

is "powerful" and strength a false equivalence here then?

[vital oracle]

theyre both words that we're using to describe asymptotic dominance

they both mean the same thing

I dont know how youre missing the english here

surely itll occur to you that if something requires multiple close-by words to attempt to describe it, its only close to but none of them?

[kindred hinge]

oh okay

im going to stare at the asymptotic dominance definition for a minute

[vital oracle]

going off of experience, youre not gaining anything from doing that

let me repeat the correct definitions

[kindred hinge]

ok

[vital oracle]

,,\lim_{x\to\infty}\frac{f(x)}{g(x)}=\pm\infty\implies f(x)\text{ dominates }g(x)
\\lim_{x\to\infty}\frac{f(x)}{g(x)}=0\implies g(x)\text{ dominates }f(x)
\\lim_{x\to\infty}\frac{f(x)}{g(x)}=c\text{ where }c\ne0\implies f(x)\text{ and }g(x)\text{ are asymptotically equivalent}

[elfin berry]

mtt

[vital oracle]

,,\lim_{x\to\infty}\frac{f(x)}{g(x)}=\pm\infty\implies f(x)=\omega(g(x))
\\lim_{x\to\infty}\frac{f(x)}{g(x)}=0\implies g(x)=o(f(x))
\\lim_{x\to\infty}\frac{f(x)}{g(x)}=c\text{ where }c\ne0\implies f(x)\sim g(x)

[kindred hinge]

I got it! (I think)

[elfin berry]

mtt

[vital oracle]

and I didnt get to keep going but the list youll usually need is:
1/x^3, 1/x^2, 1/x, 1/√x, 1, ln(x), √x, x, x^2, x^3, e^x, x!, x^x

[kindred hinge]

it literally just the limit it produces. so if you find the limit using lhopitals, and it just so happens to be +/- inf, we know that f(x) must dominate g(x)

[vital oracle]

yea

also keep in mind

[vital oracle]

by using this definition, you ignore constants changing the order

2x ~ x

2e^x ~ e^x

x = o(2e^x)

99999999x = o(2e^x)

youll notice there that $\lim_{x\to\infty}\frac{99999x}{2e^x}=\frac{99999}{2}\lim_{x\to\infty}\frac{x}{e^x}=\frac{99999}{2}\cdot0=0$

[elfin berry]

mtt

[kindred hinge]

Ok, you had me then you lost me

[vital oracle]

do you know that you can move constants out of limits

[kindred hinge]

I did not know that

[vital oracle]

bruh

maybe you need to go to khan academy and relearn calculus

[kindred hinge]

bro im in precalc

[vital oracle]

then what makes you think you can handle ∞ - ∞ if you cant handle that

you need to go one step at a time

[kindred hinge]

We were never taught anything beyond limit substitution

and specifically how to handle that case, but I didnt quite understand why it worked so it was hard to remember

[vital oracle]

then what kind of excuse is this

thats a strong accusation

limits are pure calculus

you cant say "no I wasnt being taught any calculus" as a cop-out when I tell you to learn it

[kindred hinge]

I will learn it then

I shouldnt be making excuses, though I genuinely did not know that you could do that until now

[vital oracle]

khan academy isnt the only place to learn calculus

but its a good place to start

once you learn the basics, then you can properly handle limits

[kindred hinge]

alright, thank you for helping me mtt

[vital oracle]

np

[kindred hinge]

.close

[marsh citrus]

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[marsh citrus]

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[mental hamlet]

Is multiplicative commutativity by a scalar not a property of a vector space?

like $a \times \mathbf{v} =\mathbf{v} \times a$

[elfin berry]

CrazyCuber217

[twilit grove]

I don't think I've ever seen it written like that, but I think it's true

[marsh citrus]

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[frosty bronze]

I am trying to solve (b) but I am getting zero which is definitely not correct

[marsh citrus]

@frosty bronze Has your question been resolved?

[blazing pulsar]

How are you getting 0 from that work

[marsh citrus]

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[frosty bronze]

oh nvm I got the third derivative wrong 🫠

[frosty bronze]

.close

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[livid falcon]

how do i do this question, it's Pythagoras in 3d. pls be nice and gentle im a sensitive stupid little scrawny 10th grader

[weak eagle]

3-2

[livid falcon]

[daring stump]

Use the slant height formula

[livid falcon]

wait so it'd be:
h^2 = 3^2 + 10^2
h^2 = 109
h = sqrt109
h = 10.44
???

i got the answer but idk how you get this

[gray moss]

what's the hypotenuse?

[late geode]

careful with your values and the side of the hypot

i'd recommend first drawing a cross section

[livid falcon]

oh alralr

[unreal adder]

its 6-4 and then you divide that by 2

thats the width of your triangle

aka the base

[unreal adder]

and the height is not the hypotenuse

so watch out there as well

[unreal adder]

the correct answer is in fact sqrt(99)

c^2 = a^2 + b^2

so,
c^2 - a^2 = b^2

[livid falcon]

ohhhhh i see i see

wait sorry i'm a bit confused on what numbers c, a and b are i've kinda been staring at the question and can't figure it out myself i'm a bit slow minded

[marsh citrus]

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[opaque sluice]

e

[opaque sluice]

hello

can somebody help me with this one

Im not sure what the question is looking for

for b

<@&286206848099549185>

[sick tinsel]

Make a diagram for it and you will see that a right triangle is formed with hypotenuse CT and base L root 2 /2 which is half the length of diagonal of the base square and third side being height of pyramid which is given

L is also given

[opaque sluice]

L?

oh 440?

[sick tinsel]

Length

Yeah

[opaque sluice]

like im not sure

what the diagonal is in this triangle

[sick tinsel]

The diagonal of the square base

Is the base of the right triangle

[opaque sluice]

oh damn right

so for b, am I supposed to calculate the distance of ct

[sick tinsel]

Yes it seems so

[opaque sluice]

how is the diagonal equal to ST tho

for a the answer says its 356

nvm it isnt

fucck me

[sick tinsel]

The diagonal is not ST

[opaque sluice]

yup

let me try again

holdup

@sick tinsel so with the diagonal length

how do I determine CT

[sick tinsel]

T is the top T' is point directly below T so T'T is height of the pyramid (given)

[opaque sluice]

length of diagonal is 522

so half of that is t1 no?

wait

no

cuz its 440 root 2

[sick tinsel]

How is it coming 522

It should be somewhere around 620

And half of that will be CT'

[opaque sluice]

mistake mb

[sick tinsel]

Pretty much multiply 440 by 0.707 to get length of CT'

[opaque sluice]

yup

[sick tinsel]

And now pythagoras

[opaque sluice]

i completely forgot about the 3d dimensions

gosh

tysm bro

[sick tinsel]

you are welcome

close the channel once you are finished