help-10

[vast willow]

(-4,0) is this point on the drawn line

[long plaza]

yes

[obtuse pebble]

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[timid silo]

I don't understand this step

the given proposition is that $\frac{a}{b} = \frac{c}{d}$

[warm shale]

Mushaar

[unique mural]

Hi

[timid silo]

oh wait it's pretty simple

nevermind

hi

.close

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[proper snow]

For a given (false) claim and proof, what skills do you use to determine where the proof is false?

[tardy epoch]

Pretty vague to give anything meaningful. Generally counterexamples or proof by contradiction

[proper snow]

I have a specific proof that I know is incorrect but I’m having a hard time to figure out which step exactly the proof is wrong

@tardy epoch

[timid silo]

don't ask to ask

[proper snow]

Oh I sent the question idk where it went

And A’ = A is complement and A\B = A - B @timid silo

[tardy epoch]

yea counterexample would be fine here

[proper snow]

But what step from the proof would be wrong?

I was thinking step 10

[tardy epoch]

counterexample would be easier

[proper snow]

Yes I know but I’m just trying to figure out where the proof is wrong

[proper snow]

And then I can proceed to this

[obtuse pebble]

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[obtuse pebble]

@proper snow Has your question been resolved?

[obtuse pebble]

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[timid silo]

So why does “squaring the circle” exist

[timid silo]

Basically my second question on #1154481698411642890

[tardy epoch]

what answer are you looking for exactly

[timid silo]

Like why

Nor how

[tardy epoch]

what

you just repeated your question

and ignored mine

[timid silo]

Why is a circle bigger than a square that it fits in

[timid silo]

Didnt

I used “like”

[tardy epoch]

🙄

[timid silo]

What about the answer

[tardy epoch]

.

[tardy epoch]

.

[timid silo]

.

💀

[tardy epoch]

that's different from squaring the circle

[timid silo]

I might saying it wrong too

[tardy epoch]

again, not answering my question

[timid silo]

Whats it

[tardy epoch]

from this, what do you want ?

what kind of answer

[timid silo]

Like proof

[tardy epoch]

what sort of proof

like a picture should be good enough

so draw the picture

[timid silo]

I cant rn

[tardy epoch]

then work on it when you can

[timid silo]

Why am i a fucking dumbass

Lot lot sorry for yalls effort

.closr

.closr

.close

[obtuse pebble]

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[sleek wadi]

Is this solvable for f(x)?

[sleek wadi]

Best I can get is this, but still cant integrate the left side from what I understand

If it's not solvable, is there another way to find the parametrization for x^2 where the velocity is constant?

[tardy epoch]

Looks separable

,w solve dy/dx = 1/sqrt(1+4y^2)

[warm shale]

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[tardy epoch]

Yea they give a function x in terms of y. Looks nearly impossible to invert

[sleek wadi]

darnit

[native inlet]

,w f'(x)sqrt(1-4f(x))=1

[warm shale]

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[native inlet]

W|A inverts it pretty well but I wouldn't solve what the separable gives algebraically

[sleek wadi]

wait I tried wlfram alpha before and it said it didnt know lol

Thanks for the help!

[warm canopy]

Mrfancy is a wolfie whisperer

[sleek wadi]

oh wait he forgot the squared lmao

thats why it worked

[tardy epoch]

,w f'(x)sqrt(1+4f(x))=1

[native inlet]

oh damn I did

[tardy epoch]

Well I'll be damned

[native inlet]

,w f'(x)sqrt(1-4f(x)^2)=1

[warm shale]

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[native inlet]

therree it is

[sleek wadi]

ok, cool

[native inlet]

yeah that looks nasty

[sleek wadi]

yeah im not sure if the inverse works for what im doing, but I can try lol

it's also only between -1/2 and 1/2 😭

maybe the way I went about trying to find this was complicated, so I guess ill just explain the whole problem:

[native inlet]

that's why you have the constant :)

https://www.desmos.com/calculator/uvpkdfekba

shift the constant the graph shifts

[sleek wadi]

I wanna parametrize the equation y=x^2

but with a constant speed

And that's where that equation came from lol

.close

[obtuse pebble]

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[fervent meadow]

I'm working through Shankar's textbook and am confused on how to proceed with this classical mechanics question (Ex. 2.7.8, (2)). We have been given an arbitrary point/coordinate transformation $$ q_i \rightarrow \bar{q}_i(q_1,..., q_n)$$ and we have already derived the equation $$ \bar{q}_i=\sum_j \frac{\partial q_i}{\partial\bar{q}_j}\dot{q_j}.$$ We want to show that $$ \left( \frac{\partial \dot{q}_i}{\partial{\dot{\bar{q}}}j} \right){\bar{q}} = \frac{\partial q_i}{\partial \bar{q}_j}.$$

So far I have that $$ \left( \frac{\partial \dot{q}_i}{\partial{\dot{\bar{q}}}j} \right){\bar{q}}= \left( \frac{\partial}{\partial \dot{\bar{q}}_j}\left[\sum_k \frac{\partial q_i}{\partial\bar{q}_k}\dot{qk} \right]\right){\bar{q}} $$ $$=\left( \sum_k \left[\frac{\partial q_i}{\partial\bar{q}_k}\frac{\partial \dot{\bar{q}}_k}{\partial \dot{\bar{q}}_j} + \dot{\bar{q}}_k \frac{\partial}{\partial{\dot{\bar{q}}_j}} \left( \frac{\partial q_i}{\partial\bar{q}k} \right)\right]\right){\bar{q}} $$ where I simply applied the product rule.

I am confused on how to continue, since isn't it possible that $\frac{\partial q_i}{\partial\bar{q}_k}$ can be a function of both $\bar{q}_j$ and $\dot{\bar{q}}_j$? Also, can someone please explain why it's important that we take the partial derivative with $\bar{q}$ held constant? Isn't holding $\bar{q}$ constant already implied by the partial derivative?

[warm shale]

thedude365

[obtuse pebble]

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[formal inlet]

this one asks to find the angle between 2 lines in R3

I did the dot product over the 2 magnitudes

aka this, which gives the angle

however it states im doing this wrong

i did -17/(sqrt(18)*sqrt(29))

giving me an angle of 138.0792

in the end

(this is all in degrees)

[obtuse pebble]

@formal inlet Has your question been resolved?

[formal inlet]

<@&286206848099549185>

[timid silo]

okay one moment

[formal inlet]

Take ur time

[timid silo]

ok sorry for the hold up

[timid silo]

i worked it through and ended -26/sqrt(89)*sqrt(61)

[timid silo]

ok lets work it out together again

from the beginning

we have two vectors

d1 = (-3,4,-3) + t(-1,4,-1)

and d2 = (0,0,-4) + s(3,-4,-2)

and cosx = d1•d2/(||d1|| ||d2||)

@formal inlet are you with me

[formal inlet]

yes

[timid silo]

ok now lets pick any values for s and t

[formal inlet]

yeah ive had all this info is right so far

[timid silo]

good

[timid silo]

since it'll lie on the same line regardless

what do you pick?

[formal inlet]

does it have to be 3 values each xyz?

[timid silo]

no t and s will just be a number

[formal inlet]

then we could just do 0,0,-4 and 3,-4,-2 as tghey are both on the lines right

[timid silo]

ok so you picked 1

d2 = (0,0,-4) + (3,-4,-2)

simplify this for me

[formal inlet]

3, -4, -6

[timid silo]

d1 = (-3,4,-3) + (-1,4,-1) and this too

[timid silo]

ok so d2 = (3, -4, -6)

[formal inlet]

-4, 8, -4

[timid silo]

and d1= (-4, 8, -4)

[formal inlet]

ahhh so I use these for the equation instead?

[timid silo]

now lets plugin and see what we get

[timid silo]

correct

[formal inlet]

cause I was just using s and t

ok lemme get my calc

[timid silo]

lemme do the same

[timid silo]

d1

nvm

[formal inlet]

105.15043

might have typed wrong

[timid silo]

let me verify

[timid silo]

i got 109 degrees

what was your dot product?

[formal inlet]

-20

[timid silo]

ok same

and the length of d1?

[formal inlet]

uh lemme find

isnt it magnitude?

or is that the same thing?

[timid silo]

ye

||d1||

[formal inlet]

sqrt(96)

[timid silo]

and d2

[formal inlet]

and d2 sqrt(61)

[timid silo]

i got all of those too

[formal inlet]

i still get same answer

im in degrees if that matter

[timid silo]

maybe my calculator is

[timid silo]

same

[formal inlet]

either way the program accepts the answer

[timid silo]

nic

good job

[formal inlet]

thanks for help 👍

[timid silo]

yeah its my calculator sorry

[timid silo]

yw

[formal inlet]

.close

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[timid silo]

what am i doing wrong here

[latent walrus]

it seems okay to me

[timid silo]

for some reason the stupid grader thingy thinks its wrong 😭

[latent walrus]

it might be because your lines end at -5 and 5, but i dont know if thats just the bound youre actually able to draw in

[timid silo]

i think that might be the reason since ive clicked "Get a similiar question" multiple times and its always consistently 0.67 out of 2 when it says i get it wrong

but i cant draw anything farther then -5 and 5 which is weird

[daring rock]

have you used the third draw tool under the graph? it looks like it's for drawing rays

[timid silo]

oh my god i think that was the problem 😭

https://gyazo.com/d777fb8395bd77c9b1cc0070b89c6cd0

[daring rock]

that makes sense cuz that score implies that it thinks only one part out of 3 is right

[timid silo]

yes

[daring rock]

lol @latent walrus i'm amazed you noticed that, the difference in the picture is so tiny

[timid silo]

ok thank you sm i think that was the problem

.close

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[silk sparrow]

I need some help w statistics

[silk sparrow]

I am working on transformations

of pmfs

here is the problem:

and here is my work

[obtuse pebble]

@silk sparrow Has your question been resolved?

[silk sparrow]

no

[slate zephyr]

What does pmf stand for

[gilded needle]

probability mass function

aka, p(x) = P(X = x)

[slate zephyr]

$p_Y(x) = P(\underbrace{Y}_{=X^4} = x) = P(X = \sqrt[4]{x}) = p(\sqrt[4]{x})$

[warm shale]

M8732

[obtuse pebble]

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[obtuse pebble]

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[silk sparrow]

is that correct @slate zephyr ?

[silk sparrow]

this is my work and a swer

[slate zephyr]

yes

you might want to be more clear than 1, 16, 81

[silk sparrow]

1^4 , 2^4, 3^4,

[slate zephyr]

you could say y is fourth power of integer for example

or you could say that fourth root is an integer

[silk sparrow]

is my conclusion correct that the probability of X = positive is zero

[slate zephyr]

no

[silk sparrow]

lol

Any suggestion on how I would go about doing that

[slate zephyr]

sorry for the confusion

[slate zephyr]

okay so this is right

by definition X is negative integers

but Y is positive integers

[silk sparrow]

Okay so I just need to change my possible values for Y and state that y is a fourth power if an integer

[slate zephyr]

*of

what you wrote it correct

[silk sparrow]

thanks for taking the time to look into this and help me out! god willing I hope to be able to pay it forward (someday)

[slate zephyr]

but the sequence 1, 16, 81 is not particularly clear

[silk sparrow]

gotcha!

.close

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[slate zephyr]

Why pay when you get it for free here

you are welcome though

[obtuse pebble]

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[round stag]

apparently this is incorrect idk how

[fierce lagoon]

The -100

Well

Yeah that ain't right

[shut lagoon]

You took +- once you isolated, but you have to take +- when you take the sqrt in the first place, then finish isolating

[round stag]

i still dont understand what to put as the answer, i have one submission left

[shut lagoon]

In particular, +-sqrt(A/P) = 1 + i/100 does not mean that +-(sqrt(A/P)-1) = i/100

You did the second one

But it is wrong

[round stag]

so it should be..

[shut lagoon]

It should be ... you that computes it

[round stag]

just to confirm the postive side is right but not the negative

correct?

[shut lagoon]

Yes

+- is attached to the sqrt

Not the outside

So your other one is ...

cue

[round stag]

like so?

[crude coral]

isnt that the same answer u tried

what he is trying to say is that
A/P = (1+(i/100))^2
+- sqrt(A/P) = 1+(i/100) this is correct

the +- is only with the sqrt and not with the entire answer

[round stag]

should i remove the negative sign in front of 100 in the second answer? is that what you are referring to?
sorry my english isnt good.

[obtuse pebble]

@round stag Has your question been resolved?

[round stag]

.close

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[dense crater]

I have a question again

[dense crater]

Let's persay the 39# is considered point E

Within this problem the professor actually used Position vecotr AD to compute the moment

Can I use Position AE as Position Vector?

[obtuse pebble]

@dense crater Has your question been resolved?

[dense crater]

.close

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[turbid charm]

.close

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[floral remnant]

I got here

[royal basin]

your handwriting is shit

[floral remnant]

I agree

[royal basin]

you have somehow managed to make the symbols a, 4 and 9 look identical

anyway

what is i^2?

[floral remnant]

4i^2?

[royal basin]

no, i^2 is not equal to 4i^2.

answer my question.

what is i^2?

[floral remnant]

Aaa, an algebraic expression?

[royal basin]

ok that's only technically correct.

i is supposed to be the imaginary unit here.

do you know what that is

[floral remnant]

Noooo...

[royal basin]

bruh

i^2 = -1 ??

you don't know this?

you were not taught this?

[floral remnant]

Nooo lol

[royal basin]

bruh what

how are you getting assigned this problem then

if you do not even know what it says

[floral remnant]

Haven't started school yet, trying to get ahead a bit

[floral remnant]

Wait are u talking about i like √-1?

[royal basin]

yes

[floral remnant]

I dont think that has anything to do with matrices in this problem lol

They used i as an expression im 90% sure

Okay wait lemme do that math using i as -1

Holup

[royal basin]

not i as -1

i as sqrt(-1)

[royal basin]

matrices don't inherently have to be filled with real numbers, you know.

but here this i is front and center in yours

[floral remnant]

Dunno, new to matrices lul

Okay it worked but i dunno how to articulate this

and the question hasent said anything about i being imaginary either 🤔

Thanks anyways

.close

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[timid silo]

hi i had these two problems i needed help with

[oblique nebula]

What are those problems?

[timid silo]

Sorry it wasn’t uploading

I am supposed to use dimensional analysis

I do have a reasoning

If we look at RHS, we see that dimension of [a] = [x] and Overall dimension of RHS is [a^n] or [x^n]

On the right hand side, we’ll have the dimension would be [log x] which has to be dimensionless

[a^n] = M^0 L^0 ^ T^0

So n = 0?

Is my reasoning correct?

.close

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[jaunty sluice]

Is this the correct way to prove?

[zenith spade]

This is valid, yes

[jaunty sluice]

hurray!

thanks 😄

.close

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[exotic zealot]

If f(x) is non-negative and it's sum over all its finite or countably infinite no. of x values in the domain is 1, then f(x) can serve as a probability mass function of a random variable X.

[exotic zealot]

How do you prove this? The text says that this follows immediately from discrete probability postulates but I don't see how. The converse is good with me. Just this.

[royal basin]

well like... you can just construct a random variable whose PMF is f(x), no?

[exotic zealot]

but how is that so apparent that the text just says "immediately" ?

[royal basin]

here's an example

let $f(x) = \begin{cases} 0.1 & x=2 \ 0.4 & x=7 \ 0.35 & x=11 \ 0.15 & x=21 \end{cases}$

[warm shale]

Ann

[exotic zealot]

showing that you can construct a random variable with PMF=f(x) is not so easy, so as to just skip its proof. Atleast that's what I think.

[royal basin]

the random variable X with this function as pmf
is the random variable which takes the value 2 with probability 0.1, 7 with probability 0.4, 11 with probability 0.35 and 21 with probability 0.15

that's it.

[exotic zealot]

can you please talk me through the three postulates that a probability measure/function has to satisfy in order for this to work? That is, to talk me through how this situation satisfies/we can make it satisfy the three axioms

[exotic zealot]

I do want to say that a big part of my question is clear now. Just this now.

[royal basin]

you want to go down all the way to the axioms...?

[exotic zealot]

i mean if you're saying this is a random variable, then it has to have a vaild probability measure/the probability function P. right?

[royal basin]

sure i guess

[exotic zealot]

because an r.v. is defined on the basis of a sample space and a sample space is defined along with a probability function, as far as I understand

[royal basin]

formally a random variable is a measurable function from a sample space Ω to the real number line.

so like

to give you a sample space specifically tailored to our random variable

let Ω = {a,b,c,d}, F = 2^Ω, and the probability measure defined by assigning to the points a, b, c and d the weights 0.1, 0.4, 0.35 and 0.15

[exotic zealot]

ok..

[royal basin]

does this satisfy you or do you want to spend 3 hours wading through the formal proof that all the axioms are satisfied?

it's going to be very unenlightening.

[exotic zealot]

can you state the non-negativity axiom for me..?

[royal basin]

P(E) ≥ 0 for all events E

[exotic zealot]

do you just make the P defined to have the additive axiom?

[exotic zealot]

does it really take 3 hours?

[royal basin]

yes

[royal basin]

i don't know, i did not count.

but maybe you want me to list out all 16 values of P because you won't believe me otherwise.

[exotic zealot]

but we have already put a restriction on P right? cuz, we said X has this specific PMF . so we are putting another restriction on P now?

[royal basin]

...

i claimed X has that PMF

i haven't yes proved it

[exotic zealot]

maybe

[royal basin]

P(∅) = 0
P({a}) = 0.1
P({b}) = 0.4
P({c}) = 0.35
P({d}) = 0.15
P({a,b}) = 0.5
P({a,c}) = 0.45
P({a,d}) = 0.25
P({b,c}) = 0.75
P({b,d}) = 0.55
P({c,d}) = 0.5
P({a,b,c}) = 0.85
P({a,b,d}) = 0.65
P({a,c,d}) = 0.6
P({b,c,d}) = 0.9
P({a,b,c,d}) = 1

here is the probability measure i define on my 4-point sample space

[exotic zealot]

but this does not satisfy the additive axiom

[royal basin]

you claim so?

[exotic zealot]

P{a,b} = P{a} +P{b}..

[royal basin]

yes/no question

[exotic zealot]

yeah

[royal basin]

ok, then give me a pair of events E and F, disjoint, s.t. P(E ∪ F) ≠ P(E) + P(F).

maybe i fucked up my arithmetic somewhere.

oh yes, i did.

hold on.

do you maintain your claim?

[exotic zealot]

you haven't edited it yet though..

[royal basin]

yes i have

[exotic zealot]

oh..ok

[royal basin]

do you claim that the measure P that i just wrote out in FULL doesn't satisfy additivity? Y/N

[exotic zealot]

no, I'm completely satisfied

but do you have an idea as to how you would prove this in general? I tried to but how is the proof so trivial still gets me.

[royal basin]

i mean like

[exotic zealot]

I meant to say, Yes

[royal basin]

yes as in you DO claim it DOESNT satisfy additivity???

or what

[exotic zealot]

no..

[royal basin]

bruh

[exotic zealot]

i agree

this is valid

[royal basin]

ok

so like

anyway

let x_1, x_2, ... be the set of values at which f(x) > 0. by assumption this set is at most countable and so can be viewed as a sequence, whether finite or infinite.

[exotic zealot]

especially they say it immediately follows from the postulates

[exotic zealot]

ok..

[royal basin]

construct a sample space Ω of as many points as needed

assert P({x_n}) = f(x_n)

then extend P by additivity

since $\sum_n f(x_n) = 1$ you will get $P(\Omega) = 1$ just fine

[warm shale]

Ann

[exotic zealot]

you mean P(X=x_n)=f(x_n)

[royal basin]

there's no X yet.

[exotic zealot]

but P is a set function..

[royal basin]

i mean you can make the points in the sample space formally different from the numbers x_n

[royal basin]

i mean {x_n} as in the event consisting of the single point x_n

[exotic zealot]

ok so you mean P{w_n} = f(x_n)?

[royal basin]

sure.

[exotic zealot]

ok..

[royal basin]

since you wanted to correct me so badly

on an inane and trivial point, i must add

but like

ok fine

P({w_n}) = f(x_n)

and then extend that by additivity to all subsets of Ω

and then X : Ω -> R will be defined by w_n ↦ x_n and it will have the required pmf

[exotic zealot]

so the additive extension guarantees that any subset will also probabilistically be non-negative , I guess?

[royal basin]

the probability of any subset will also be nonnegative

[exotic zealot]

on a personal note, do you really think this follows immediately though?

my doubt is cleared btw. Good explanation i must say

[royal basin]

yes

because all of these constructions are... less intimidating than the notation might make them appear

[exotic zealot]

oh..have you seen other texts like this where they say immediately follows, but is even more not-so immediate than this?

do they get progressively more "skip obvious steps" and along with that, do you also get used to it?

[royal basin]

i guess so

[exotic zealot]

ok. thank you.

.close

[obtuse pebble]

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[bright geyser]

Yo i’m stuck trying to understand besides identity

[bright geyser]

bedouts

besouts

BESOUTS

BEZOUTS

this one

I understand that d will divide ax + by

but not that d = ax + by

would love some help

[royal basin]

there EXIST x and y such that ax+by is exactly d

if you want, give me two integers and i will tell you their GCD and a bézout expression for it

[hasty salmon]

Well that’s the whole point of bezouts identity

You can look at the proof

[bright geyser]

well ye

[bright geyser]

i’ve looked at a bunch of proofs i don’t get them

i just coke across stuff like this that seemingly uses the proof to prove itself

[bright geyser]

yaa but why

it would just be helpful to work through a proof with someone

[royal basin]

brb

[bright geyser]

ight

[royal basin]

ok im back

so like do you want to work through a pre-written proof or do you want a demonstration with an example

[bright geyser]

i think this would be better 🙂

[royal basin]

"this"?

which one would you like tho

[bright geyser]

whoops sorry

second one

demonstration with example

[royal basin]

ok

give me two integers

any two, but preferably ones not equal to each other nor zero

[bright geyser]

11 and 64

[royal basin]

ok, let's see...

gcd(64, 11) = 1 = 64 * 6 + 11 * (-35)

,calc 64 * 6 + 11 * (-35)

[warm shale]

Result:

-1

[royal basin]

oops

flip that

64 * (-6) + 11 * 35

is 1

[bright geyser]

hmm ok

maybe a full proof would help acyually

[royal basin]

do you have a proof that you're reading somewhere?

[bright geyser]

i haven’t been given one formally no

[royal basin]

can you share at least one of the proofs you looked at

[bright geyser]

ye sorry i’ll just be a few minutes

[royal basin]

ok

ping me once you can show me the proof

[obtuse pebble]

@bright geyser Has your question been resolved?

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[royal basin]

.reopen

[obtuse pebble]

✅

[royal basin]

@bright geyser ya still there or what

[bright geyser]

sorryy, I think I actually found a proof

this one was a little strange but it did make sense to me

[obtuse pebble]

@bright geyser Has your question been resolved?

[restive stratus]

how to count the cos?

i forget

[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[restive stratus]

no just cos

[tranquil yarrow]

?

[restive stratus]

in a triangle

[tranquil yarrow]

what?

[royal basin]

!occupied

[obtuse pebble]

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

[royal basin]

!status

[obtuse pebble]

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

[tranquil yarrow]

2

[keen garnet]

!show

[obtuse pebble]

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

[restive stratus]

forgot about it

[tranquil yarrow]

i need to prove that the initial equation has no roots
and/or that my polynomial equation has no root in the interval [0; 1]

[restive stratus]

i ment the simple cos

[royal basin]

you took t = sin^2(x) yes?

[tranquil yarrow]

yeah

[royal basin]

so that means t ranges from 0 to 1

[tranquil yarrow]

yes

[royal basin]

and from the graph you see the function has no zeros between 0 and 1

[tranquil yarrow]

exactly

[restive stratus]

how old are you

[tranquil yarrow]

but i cannot graph this on a piece of paper because it's too complicated

[tranquil yarrow]

me? why does that matter?

[restive stratus]

say it

come on

[royal basin]

are you going to help

or are you just gonna keep interrupting

bc so far you've said nothing that was helpful or relevant to the question at handd

and OP's age is not relevant either

[restive stratus]

i am 13 you genius

how can i help you

[royal basin]

good for you but nobody asked your age

[keen garnet]

who asked

[restive stratus]

nnnnnn

[royal basin]

anyway ok so like

i need to prove that the initial equation has no roots
hm

[tranquil yarrow]

yeah

[royal basin]

can we come up w/ some clever bounds here

cos^6(x) + sin^8(x)

what can we do to this

[tranquil yarrow]

turn cos^2x into 1-sin^2x and vice versa

idk really

[royal basin]

wait

do you have access to double-angle identities and such

[tranquil yarrow]

ofc but why do i need them

the only thing i've come up with is to find the derivative, then determine the point where f(x) is the smallest

[royal basin]

overkill imo

[tranquil yarrow]

but as i said it didn't help due to me not being able to solve the cubic equation f'(t)=0

[royal basin]

oh this is EASY actually

[tranquil yarrow]

what do you suggest?

[royal basin]

$\sin^8(x) \leq \sin^2(x)$ and $\cos^6(x) \leq \cos^2(x)$

[warm shale]

Ann

[royal basin]

so lhs ≤ 1

[tranquil yarrow]

uhmmm interesting

but it doesn't prove anything

as far as i can see

i mean like

i understand what you mean but mathematically for this to mean something sin(x) and cos(x) have to be equally monotonic

[royal basin]

???

you don't need any monotony for shit

these two inequalities imply that FOR ALL REAL $x$ we have $$\sin^8(x) + \cos^6(x) \leq \sin^2(x) + \cos^2(x) = 1$$

[warm shale]

Ann

[royal basin]

therefore $\sin^8(x) + \cos^6(x) \leq 1$ for all $x$

[warm shale]

Ann

[royal basin]

but 5/4 is greater than 1

do you agree or disagree? @tranquil yarrow

[fervent cradle]

oh lol that's funny

trick question

[tranquil yarrow]

i agree with what you're trying to say but disagree with your conclusion

[keen garnet]

1>5/4 ?

[tranquil yarrow]

not that

[royal basin]

can you tell me what you disagree with

like quote me

[tranquil yarrow]

this

i know it's true but that's definitely not how you prove it

[royal basin]

for all real numbers $a, a', b, b'$ if $a \leq a'$ and $b \leq b'$ then $a+b \leq a'+b'$

[warm shale]

Ann

[royal basin]

agree or disagree?

[tranquil yarrow]

I agree but sin(x) and cos(x) are not constant numbers, they're functions

for every x in [0; 2pi] they vary

[royal basin]

this reasoning applies to them all the same just with a ∀x ∈ R stuck to the front

[royal basin]

and the inequality is true pointwise

[keen garnet]

can you find a value of x when $sin^8(x)>sin^2(x)$

[warm shale]

Joseph.P

[tranquil yarrow]

okay Ann i have a question for you

[royal basin]

hit me

[tranquil yarrow]

how do you prove $$2cos(x) + 5sin(x) = 8$$ has no solutions

[warm shale]

exterminated

[royal basin]

$|2 \cos(x) + 5 \sin(x)| \leq 2 |{\cos(x)}| + 5 |{\sin(x)}| \leq 2 \cdot 1 + 5 \cdot 1 = 7$

apply triangle inequality

[warm shale]

Ann

[royal basin]

or even simpler

[keen garnet]

or Caushy Swartz inequality

[tranquil yarrow]

its not because max(cos(x)) == max(cos(x)) = 1 right?

[royal basin]

what

[tranquil yarrow]

bro i'm in high school

[royal basin]

i never spoke about maximum

[tranquil yarrow]

uhm okay

[royal basin]

sorry, can you restate your question to me

because it looks like you take issue with the proof i just presented but i do not know what your issue is

[tranquil yarrow]

yeah well okay

i agree

gj

[obtuse pebble]

@tranquil yarrow Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[formal heron]

what.. im lost and idk how to do any of those..

[formal heron]

Be it a , b > 0 . Demonstrate the unequals ?

the stuff inside () is just a few tips in order to help us

but idk anyhting

[static marten]

with the first question, how could you make the inequality easier to prove

[formal heron]

well i could multiply the 2 with the a2+b2 and the 1 with the shorted formula that i got below ?

or idk

[static marten]

which formula?

[formal heron]

since i could probably make the a2 + b2 into this form

[formal heron]

uhm a^2 +_ 2ab + b^2

[static marten]

$2(a^2+b^2)\geq a^2+2ab+b^2$

[warm shale]

chlamydia

[static marten]

is this what you mean?

[formal heron]

yeah..

[static marten]

right, so where can we progress from here?
having terms on both sides of the inequality is a bit inconvenient

[formal heron]

sorry but what do you mean by terms ? since english is my second language i dont really understand a lot of stuff

[static marten]

all the a and b

[formal heron]

ah

[static marten]

i'll be careful

[formal heron]

we could make it 2a^2 + 2b^2 and then we could substract ? ( - ) a2 and b2

but idk if it'll head us towards a dead end or not

[static marten]

sure, so that gives us

$a^2+b^2-2ab\geq0$

[warm shale]

chlamydia

[static marten]

does this look familiar?

[formal heron]

i mean it does..

so we just redo the formula said previously?

and we end up with (a-b)^2 ?

[static marten]

$(a-b)^2\geq0$

[warm shale]

chlamydia

[static marten]

and because it's squared, it's always true

[formal heron]

wait that was it ?

[static marten]

what were you confused about

for a,b>0, that's all

[formal heron]

wait but how did we substarct the a and b from this and got this outcome ?

or

nvm

nvm

we moved the 2ab as well in that process

with the changed symbol

[static marten]

yeah

it works the same way as an equality, always change sign when moving it

[formal heron]

so what about doing 2 now ?

[static marten]

so what's the obvious problem in 2?

[formal heron]

so we could ^2 it all

or

[static marten]

sure

[formal heron]

ye that will probably be the way

[static marten]

$\frac{a^2+b^2}2\geq\frac{(a+b)^2}4$

[warm shale]

chlamydia

[static marten]

$2(a^2+b^2)\geq a^2+2ab+b^2$

oops

[warm shale]

chlamydia

[static marten]

you can see where i got this?

[formal heron]

we moved the 4 to the other side by multiplication

right ?

[static marten]

yeah, and expand (a+b)^2

[formal heron]

yeah

[static marten]

and now...

[formal heron]

now it is just like last problem.

a^2 + b^2 - 2ab >_ 0

i think

$a^2+b^2 - 2ab\geq 0

[unique mural]

Hi

[static marten]

$a^2+b^2 - 2ab\geq 0$

[warm shale]

chlamydia

[formal heron]

so this is also right ?

[unique mural]

Would you be free to help me out with a question

[static marten]

yeah

[formal heron]

alright , thanks

[static marten]

typically you'd need to go to #❓how-to-get-help

[formal heron]

the teacher gave a tip of " at end 2 can be given as a common factor"

what does that mean..

[static marten]

maybe it's talking about multiplying the 2 to simplify?

[formal heron]

probably

but we didnt need it

so it's alright

this is where i tried and failed so hard

but i'll try

[static marten]

did you square both sides

[formal heron]

ye

i squared it all

[static marten]

and you get a nasty right side

but because you have ab on both sides, you can do something with that first

[formal heron]

idk what we can do.. the only thing i thought of is multiplying both sides by (a+b)^2 to get rid of that from the right side

[static marten]

sure, but we can divide by ab too

[formal heron]

yeah that also but idk how to do that

[static marten]

$(a+b)^2\geq ab$

[warm shale]

chlamydia

[static marten]

why not

[formal heron]

since dividing was an option.. but how do we divide the terms?

i've never done this so idk how it's done

how do i divide 2a to a ? it'll give 2 ?

[static marten]

$ab\geq \frac{(ab)^2}{(a+b)^2}$

[warm shale]

chlamydia

[static marten]

what 2a

[formal heron]

it was an example

[static marten]

oh yeah

[formal heron]

dividing it all by ab ?

[static marten]

yes

[formal heron]

that's plausible but how do i do it tho

ab divided by ab = ?

[static marten]

because a,b>0, you can multiply or divide with no problem

[static marten]

when you divide 4 by 4, what do you get?

[formal heron]

1

so i just write the result in numbers?

[static marten]

you are dividing (ab)^2 by ab here

[formal heron]

that's where i had troubles.

[static marten]

what do you mean?

[formal heron]

i dont understand how i can divide them since they aren't numbers

i know the principle but idk the way

[static marten]

i wouldn't say they aren't numbers, it's like they're pretending to be numbers

so ab represents a number

[formal heron]

but it is squared

[static marten]

yes

so you have a number squared divided by the number

[formal heron]

so do we get the root of that number?

the root of it**

like sqr ab ?

[static marten]

ok think of it this way

(ab)^2 can be expressed as (ab)*(ab)

right?

[formal heron]

yeah

[static marten]

so when you divide that by ab

[formal heron]

we get 1x1 ..

or?

[static marten]

what is this

[formal heron]

1*1

[static marten]

you only divide by ab once

[formal heron]

2?

[static marten]

like if you have 7^2 divided by 7

[formal heron]

it's 7

[static marten]

now pretend that ab=7

so (ab)^2 divided by ab

[formal heron]

so we get ab?

[static marten]

yes

[formal heron]

$ab\geq \frac{(ab)^2}{(a+b)^2}$

so we were here

[warm shale]

General.Admiral MWstudios

[formal heron]

so 1 >_ ?

[static marten]

we'll multiply the (a+b)^2 too

$(a+b)^2\geq ab$

[warm shale]

chlamydia

[static marten]

right?

[formal heron]

so we multiply it all by (a+b)^2

wait how did we go from a*b to (a+b)^2

oh

[static marten]

$1\geq\frac{ab}{(a+b)^2}$

from before

[warm shale]

chlamydia

[formal heron]

yeah

[static marten]

yup

[formal heron]

$(a+b)^2\geq ab$

[warm shale]

General.Admiral MWstudios

[formal heron]

then we simply move it to the left

[static marten]

i'll tell you now, you can't use the same trick

[formal heron]

do we open it ?

then move it ?

[static marten]

$a^2+ab+b^2\geq0$

[warm shale]

chlamydia

[static marten]

nvm you can

because a,b>0, so all of it >0

[formal heron]

wait but

what happened to that ab

oh

nvm

nvm

[static marten]

it's ok

[formal heron]

so we simply - it from the expanded left side

since it was positive , we extracted it from the left row

right?

[static marten]

yes

[formal heron]

and we cant do anything with this

so we simply leave it as it is?

i mean we got what we wanted

[static marten]

yeah, there's no good way to factorise it

[static marten]

i think this is good

[formal heron]

yeah

[formal heron]

well

[static marten]

same problem of square root

$(a+2)^2\geq2^2(a+1)$

[warm shale]

chlamydia

[formal heron]

so you but it all on ()^2 ?

[static marten]

yes

[formal heron]

so we get (a+2)^2 which is already a formula and it'll surely be used . on top of a+1

so it'll remain the same?

like we remove the a+1?

[static marten]

what do you mean remove

subtract?

[formal heron]

idk how we got rid of it

i mean we got (a+2)^2 on top from the ()^2 . but on the lower part we got a+1

[static marten]

$\frac{a+2}{\sqrt{a+1}}\geq2 \Rightarrow (a+2)\geq2\sqrt{a+1}$

[warm shale]

chlamydia

[static marten]

$(a+2)^2\geq 2^2(a+1)$

[warm shale]

chlamydia

[formal heron]

ohh so we moved itt

this is the first time i see someone move from the left to the right

[static marten]

it's not about left or right, it's about different things that make work hard, like fractions or square roots

[formal heron]

alright

it does make sense now

so now we simply open the left side

a^2+2ab+4 >_ 2^2(a+1)

[static marten]

woah

expand (a+2)^2 again

[formal heron]

i mean the formula is (a+b)^2 = a^2 + 2ab + b^2

[static marten]

yes

[formal heron]

so replacing that will be

[static marten]

and this time, b=2

[formal heron]

(a+2)^2 = a^2 + 2a2 + 2^2

OH DAMN

I put b

instead of 2

i just noticed it

[static marten]

oh lol

[formal heron]

so

a^2+ 4a + 4 ?

[static marten]

yeah

and the right side is?

[formal heron]

$a^2+4a+4\geq 2^2(a+1)$

[warm shale]

General.Admiral MWstudios

[formal heron]

so then..

wait

it gives us 4* a+1

so 4a + 4

[static marten]

yup

[formal heron]

so it gives us $a^2\geq 0$

[warm shale]

General.Admiral MWstudios

[formal heron]

right?

[static marten]

yes

[formal heron]

wth is that A.

[static marten]

this one is special, because you have a pattern

[formal heron]

yeah..

[static marten]

the upside down a?

[formal heron]

yeah

[static marten]

it means for all

[formal heron]

ah mb

[static marten]

$(t-\frac1t)^2\geq0$

[warm shale]

chlamydia

[static marten]

i want you to think about this

and see if you can get something nice out of it first

[formal heron]

i need to compare both of them

[static marten]

not yet

this is separate

[formal heron]

ah

[static marten]

you good?

[formal heron]

im stuck

i mean

[static marten]

did you expand the left?

[formal heron]

multiply it by t ?

..

no

OH DAMN

[static marten]

ah

[formal heron]

that's right

we could use that

t^2 - 2 * t * 1/t + t^2 >_ 0

right?

[static marten]

yeah

so about the 2t*1/t

that becomes 2, right?

[formal heron]

$2t*\frac1t$

[warm shale]

General.Admiral MWstudios

[formal heron]

i mean

wait 30 secs for me to process this

[static marten]

take your time

[formal heron]

yeah that's right

since if we replace t with any number other than 0

i mean greater than 0

it'll give us 2

for example 3

[static marten]

yeah, because you have t and you're dividing it away

[formal heron]

2* 3 multiplied by 1 and 3 it is 6 mult. 1/3 is 2

it's ilogical and logical at the same time

[static marten]

lol

so we have $t^2+\frac1{t^2}-2\geq0$

[warm shale]

chlamydia

[static marten]

$t^2+\frac1{t^2}\geq2$

[warm shale]

chlamydia

[formal heron]

so it is t+ 1 ?

or t^2 + 1

[static marten]

what do you mean

[formal heron]

dont we uhm

like add the t^2 to the 1/t^2

[static marten]

oh, don't worry about that

[formal heron]

and get t^2 + 1 / t^2

[static marten]

this is the big part i wanted you to see

if we look back at the question

[static marten]

$(x+1)+\frac1{x+1}\geq2$

[formal heron]

..

[static marten]

do you see something similar?

[warm shale]

chlamydia

[formal heron]

damn.

[static marten]

this tells us that anything+1\anything>=2

for positive numbers

[formal heron]

yeah..

[formal heron]

...

damn

so (x+1)+1 / x+1

it is 1+1 ?

so 2..

[static marten]

no, it's because of this being true

[formal heron]

x

[static marten]

so we use that to make something useful

are you struggling?

[formal heron]

a bit..

since i havent done math in 3 months and i just started highschoolk

[formal heron]

can try ()^2 ?

or multiply it all by x+1

[static marten]

multiplying would be better

it should still work out

[formal heron]

yeah it'll be easier..

[static marten]

yeah maybe it wasn't good to try to teach that

[formal heron]

what?

try to teach what

[static marten]

this

we could've just done multiplying

but that comes in useful for harder questions

[formal heron]

i'll end up using it as well since im on math and comp science profile..

$(x+1)+\frac1{x+1}\geq2 | * (x+1)$

so we end up with

nvm

multiplying mbb

[warm shale]

General.Admiral MWstudios

[formal heron]

$(x+1)^2+\frac1(x+1){(x+1)^2}\geq2*(x+1)$

or.

wait

[static marten]

about the middle term

[formal heron]

i noticed it as well

[warm shale]

General.Admiral MWstudios

[formal heron]

wait wtf

[static marten]

use {(x+1)}

$(x+1)^2+\frac1{x+1}(x+1)\geq2(x+1)$

[warm shale]

chlamydia

[static marten]

right?

[formal heron]

ah yes

[static marten]

so you have x+1, and you're dividing by itself

[formal heron]

so we get 2 ?

[static marten]

please think carefully about this, if you have a number, and you divide by the same number...

[formal heron]

we get 1 ?

[static marten]

yeah

[formal heron]

nvm

we multiplied the 1 with the x+1 ?

[covert kettle]

pls where can i find help 0

[formal heron]

so we got x+1 / x+1?

[formal heron]

#❓how-to-get-help

[static marten]

yeah

[formal heron]

we multiplied only the upper part

[static marten]

she already has a channel open lol

[formal heron]

that's alright

[formal heron]

right?

[formal heron]

right**