help-17

[vapid wedge]

no?

we will have asinax/ax

[livid horizon]

Yeah

But a is a fonstant

Constant

Ohhh

[vapid wedge]

If we have a in nominator, we can't use the formula

[livid horizon]

I mean that you'll still get a term sinax/ax, not just that

We can

If there's a constant multiplied within a limit

Then we can multiply it with the whole thing

[vapid wedge]

$\frac{asin(ax)}{ax}*\frac{1}{x^2+1}$

[twin meteor]

Merineth

[vapid wedge]

So we have this?

[livid horizon]

Yes

[vapid wedge]

oh and now combine back together?

[livid horizon]

Now if you put the limit x to 0

sinax/ax becomes 1

[vapid wedge]

no

it becomes 0

[livid horizon]

And the other thing becomes 1/2

[vapid wedge]

sin(0) = 0

and the denominator beceoms zero

[livid horizon]

But lim (sinx)/x is 1

[vapid wedge]

can't divide by zero

[livid horizon]

.

[vapid wedge]

ooooh

so you mean

$\frac{a}{a}\frac{sin(ax)}{x}\frac{1}{x^2+1}$

[livid horizon]

Hence, lim sin(ax)/ax = 1

[twin meteor]

Merineth

[livid horizon]

Yeah, you can take sin(ax)/ax as one term

[vapid wedge]

so the middle one is = 1

right one = 1

[livid horizon]

So it's (asinx)/(ax(x²+1)

[livid horizon]

Ye

[vapid wedge]

what about a/a?

[livid horizon]

And a is as it is(in the numerator)

[vapid wedge]

but then we get

a*1

[livid horizon]

Yes

[vapid wedge]

But don't we want to know what a is?

[livid horizon]

If the value of a isn't stated then we can just leave it as it is

[vapid wedge]

But i want to determine a

if lim x-> 0+

[livid horizon]

a is a constant

[vapid wedge]

Now we only found out that a*1

Yes

but what constant

[livid horizon]

Is that the entire q?

[vapid wedge]

No.

I want to determine a so the y = 1

[livid horizon]

What's y?

The limit?

[vapid wedge]

Decide a and b so it becomes continuous in x = 0

I already figured out the bottom function to be y = 1

[livid horizon]

Ohhh

[vapid wedge]

and now i need to determine what a should be for it to also pass y = 1

[livid horizon]

I see

Ok so

We got the limit to be a

And we also want it to be equal to 1

[vapid wedge]

Yea

[livid horizon]

Just set both equal

[vapid wedge]

so a = 1

[livid horizon]

Yes

[vapid wedge]

Ah that makes sense

meaning B is also 1

so answer: a=1, b=1

[livid horizon]

Yes

[vapid wedge]

pog

tysm!

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[vapid wedge]

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[bleak crow]

Does f(x)=f(2x)=f(2x+1) mean that f is constant

[bleak crow]

For a function defined on positive integers

[lyric relic]

no

[lime basin]

no

no

[bleak crow]

I just edited it

[urban edge]

Still no

[bleak crow]

I also have that f(x)=f(2x)

[urban edge]

Wait

[safe sigil]

no

[lyric relic]

let $f=\begin{cases} f(2n)=n\f(2n+1)=n\end{cases}$

[twin meteor]

everg

[covert perch]

bro

help me too

wtf

#help-4

[lyric relic]

f(4)=2 but f(6)=3

[median crane]

If you impose continuity then it has to be constant, otherwise…

[bleak crow]

Well in this case I also know that f(x)=f(3x)

sorry for not mentioning it I didn’t know it would make a difference

[lyric relic]

[hard atlas]

well also f(4)=f(2)=f(1)=f(3)=f(6). so that counterexample doesnt work

[bleak crow]

Yeah true

So f(3x)=f(x) doesn’t even matter here

But is there a way to prove it?

If f(x)=f(2x)=f(2x+1), how to prove that f is constant on integers

[hard atlas]

induction

prove that f(n)=f(1)

[lyric relic]

so is f(x)=f(2x)=f(2x+1)=f(3x)?

[bleak crow]

Yes

[hard atlas]

wait it isnt constant actually

[bleak crow]

Is there a counterexample?

[hard atlas]

if n is positive, then f(n)=0 but if n is negative then f(n)=f(-1)

[bleak crow]

Yeah but this is for positive integers

[hard atlas]

ok sry just read this message

[lyric relic]

it is

[bleak crow]

Yeah I’m sorry I wrote it here but forgot to mention it later

[hard atlas]

then yeah, like I said. induction

[bleak crow]

so basically suppose f(x)=f(1) and prove f(x+1)=f(1)

[lyric relic]

strong induction it's perfect here

[bleak crow]

Alright, thanks everyone!

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[vast shale]

someone help

[lime basin]

with?

[vocal sleet]

@vast shale Has your question been resolved?

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[sleek sluice]

ok so i reparamaratized everything with respect to T so i get <t,t^2/2,t^3/6>

that is x=y^2/2 and zy^3/6

so i derived that to get <1,t,t^2/2>

took the absolute value to get (t^4/4+t^2+1)^(1/2) which can be factored to (1/2)(t^2+2) or t^2+1 and put it in an integral from 0 to 8

but i got 280/3

[gaunt sparrow]

Wouldn't your parametrization induce y = x^2/2 ?

[sleek sluice]

oh so it shouldnt be t^2/2

but instead since y= (2x)^(1/2) then we get (2t)^(1/2)

same with z then

[gaunt sparrow]

To get rid of sqrts you can always make x = t^2

[sleek sluice]

and z=((2x)^(3/2))/6

thats true

[gaunt sparrow]

It'll make the arclength easier to compute

[sleek sluice]

okay i'll rewrrite my answer

[vocal sleet]

@sleek sluice Has your question been resolved?

[sleek sluice]

so its a bit messy but im still not getting a nice number, am i still making a mistake with the paramatization

[gaunt sparrow]

Last line should be $\sqrt{2}\sqrt{t^4 + 2t^2 + 1}$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

And the inside is a perfect square.

[sleek sluice]

oh fuck 😭

okay, that makes sense

thanks again

[gaunt sparrow]

No worries

[sleek sluice]

sorry for bothering again, but even with that perfect square, i still wouldnt get 44/3?because then we have an integeral from 0 to 8 of sqrt(2)(t^2+1) would is not 44/3?

[bronze osprey]

you did t^2 = 8

so you should be integrating from x = 0 to sqrt(8) instead

[gaunt sparrow]

Yeah gotta change the bounds.

[bronze osprey]

yep that does indeed give 44/3

[sleek sluice]

ah so t=sqrt8 then?

then that gives the proper answer

[bronze osprey]

ye

[sleek sluice]

okay thanks alot!!

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[lunar ice]

alguien sabe de micro?

[lunar ice]

microecomia

i need hepl with this

[vocal sleet]

@lunar ice Has your question been resolved?

[flat whale]

You should translate if you want someone to help

[lunar ice]

Consider the case of a monopolist whose cost function is c(q) = q^2 and whose demand function
The inverse is given by p(q) = a – q.

1. Pose the maximization problem and solve it (prices, quantities, profit).
2. Suppose that the monopoly can make a single payment of K pesos in advertising in such a way that
   the inverse demand is p(q) = 2a–2q (that is, it is a less elastic demand). Find the maximum value
   of K that the monopoly is willing to pay.

[rapid swift]

!status

[vocal sleet]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vocal sleet]

@lunar ice Has your question been resolved?

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[ruby brook]

ok so im on part D

idk what to do text

next

<@&286206848099549185>

[summer lichen]

Try using u-sub

[ruby brook]

it aint helpin

im getting nowhere

[summer lichen]

hmm aight

lemme solve it fully first

[timid plank]

amk cocu Einsteinmisin

[summer lichen]

yeah you can do it via u-sub

[ruby brook]

ikik i did it wrong

i did it again and im close but im getting 7/2

the answer key says -7/2

that would be 7/2

hm what did i do wrong

ah i see it

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[summer lichen]

you put two negatives

[ruby brook]

nah i switched f(4) and f(1)

f(4) is 10

my negatives r hella messy

thats prob why

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[robust heron]

could someone help me out with c and d??

i think i have a and b right but im not exactly sure how to continue

i know the process to find the marginal pdf here, but im not certain on the bounds of integration here

[vocal sleet]

@robust heron Has your question been resolved?

[robust heron]

<@&286206848099549185> (:

[steel gulch]

I would help but I don't even understand this 😭 I think I should remove myself from helpers cuz I'm no helper I'm in 8th grade lol

[robust heron]

no worries lol

[steel gulch]

<@&286206848099549185>

[sweet flower]

Your b says the probability that x+y<0 is greater than 1

[robust heron]

i divide afterwards

that was just the area in blue (just poor notation on my end)

[sweet flower]

Also notice ‘divide by total area’ is the same as ‘multiply by c’

[robust heron]

right

ive seen that in another example as well

but i just cant understand which bounds to use here

because im going to have to integrate with respect to y

[sweet flower]

Yep that’s how you get the marginal for x

[robust heron]

yeah

im just not certain on which bounds to use

i was thinking rearrange for y?

but then i couldnt get a lower bound

[sweet flower]

Well you can use that, I think cylindrical coordinates look easier here

[sweet flower]

You would need 2 integrals

Or double the area of the upper half circle

[robust heron]

this is the example im trying to referencce

i understand how he came up with x to 1 as that bound

but i dont see how i can get mine (using a similar idea) here

[sweet flower]

So you know you’ll have to do a trig sub somewhere since the result is 1/pi

[vocal sleet]

@robust heron Has your question been resolved?

[sweet flower]

So the bounds you could use are -1<=x<=1 and 0<=y<=sqrt(1-x^2)

[robust heron]

oh wait

that makes sensee

however i dont see why its 0 on the lower bound of y

and not -1

actually that makes sensee

its just calculating a different part of the circle, but same size as what i drew

[sweet flower]

It’s the top half

[robust heron]

yeah

ok that makes sense

thanks so much

[sweet flower]

Then 2* that is how you would find A without using the area of a circle

[robust heron]

right

[sweet flower]

And you can use that same method for C

[robust heron]

yeah i figured so, but they ask us to use the method where you verify the pdf=1

[sweet flower]

But with xp(x) instead of p(x)

[robust heron]

ok yeah i think i have read that somewhere

[sweet flower]

For A or C?

[robust heron]

C

oh, part A, solving for c

[sweet flower]

Ah okay yeah

[robust heron]

anyways thank you, that question had me stuck for a bit

[sweet flower]

Did you get parts c and d?

[robust heron]

well i havent done it yet, but im going to end up doing it later tonight lol

i just couldnt get by the bounds

so kinda stopped there for now haha

but as for the process yeah i should know how to do it now

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[robust heron]

could someone help me out on this question

not sure how to start exactly

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[knotty shell]

is this (.6,0)?

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@knotty shell Has your question been resolved?

[knotty shell]

<@&286206848099549185>

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@knotty shell Has your question been resolved?

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@knotty shell Has your question been resolved?

[flat whale]

!show

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[winter kestrel]

Hello, here isn't 2y +2x supposed to be bracketed? why is dy/dx only applied to 2x

[paper depot]

(2xy)' = (2x)' * y + 2x * y'

so no, the 2y+2x is not supposed to be bracketed

[winter kestrel]

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[tired saddle]

Hi, I need a little help with this sequence question, I have gotten a bit through it but got stuck

[tired saddle]

i'll provide some working

this is where i got stuck

[livid horizon]

$S_n = \frac{n{(2u_1 + d{(n-1))}}}{2}$

[tired saddle]

i forgot to put the u_1 for some bits of the working sorry

[twin meteor]

Lorentz

[lapis marten]

You can substitute u1 = 6 - 3d

In the bottom equation

[livid horizon]

Also u4 = 6 = u1 + 3d, you can write u1 in terms of d and substitute in the other equation (s)

[tired saddle]

omg

[livid horizon]

Ah yes, didn't see you typed that mb

[lapis marten]

dw

[tired saddle]

thank you guys so much

luv you

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[muted brook]

Not even sure where to start with these, any help greatly appreciated.

[vast shale]

what does it mean if a function is one to one

what test must it pass

that will be the key to answer the first few questions

[muted brook]

but how do i find the inverse?

[vast shale]

the way i do it is

swap the x and y

and rearrange to make it y=...

eg i want to find inverse of y = x + 2
swap x and y: x = y + 2
rearrange: y = x - 2

[muted brook]

thank you👍🏾

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[digital roost]

CAN SOMEONE EXPLAIN ME HOW THEY CAME UP THAT LONG SOLUTION

[digital roost]

I MEAN ABOUT THAT 3 STEPS: MULTIPLYING ON BOTH SIDES BY y

[pale perch]

WOW PROBABLY BECAUSE THEY COMBINED THE FRACTIONS

[digital roost]

i dont know where that simplification came from can someone explain me step by steps

[digital roost]

i do that but my answer didnt much

[pale perch]

how so, show it

[digital roost]

or i just make it wrong

[hardy vector]

so the goal is too find dy/dx. You could have just applied quotient or product rule.

[vocal sleet]

@digital roost Has your question been resolved?

[heavy yoke]

when we use logarithmic differentiation, we always "could have" done it purely with product, chain, power, quotient rules, etc. but that would turn out to be massively more complicated than this way. even though this isn't simple it's still simpler than the alternative

[digital roost]

Then can you tell how where that long solution came from? or what method use it im struggling to find way i want to know and learn it badly

[heavy yoke]

https://tutorial.math.lamar.edu/Classes/CalcI/LogDiff.aspx

the general rule is just to take the natural log of both sides, split into different logarithms, use implicit differentiation, then solve for dy/dx

[digital roost]

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[dim path]

how did it become cos x ?

[livid horizon]

lim sin(h/2)/(h/2) as h tends to 0 = 1

[tidal dock]

$\lim_{h\to 0} \frac{\sin(h/2)}{h/2} = 1$

[twin meteor]

artemetra

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@dim path Has your question been resolved?

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[urban laurel]

i got an elementary question

[urban laurel]

$\sqrt{AB} = \sqrt A \times \sqrt B$

[twin meteor]

Adam Chebil

[vast shale]

Shoot

[urban laurel]

here, i can only split the sqrt if A and B are positive right ?

[livid horizon]

Either one can be negative too

[urban laurel]

so if A>0 and B<0 i can split ?

[vast shale]

we can split the square root of a product if both A and B are positive numbers.

[livid horizon]

Yes like
sqrt(-3) = sqrt(3) * sqrt(-1) = isqrt3

But both can't be negative

[hybrid flicker]

If A>0 and B<0 then sqrt(AB) isn't well defined in the first place

[urban laurel]

prof doesn't accept negative number under sqrt

[livid horizon]

Oh

[vast shale]

We cannot

[livid horizon]

Then it's just positive, yes

[urban laurel]

so the only way to split is when A and B are both positive ?

[vast shale]

we cannot split the square root of (AB) in the same way if either A or B is negative

only when BOTH is positive

[livid horizon]

Why not

[livid horizon]

Is this wrong then?

[urban laurel]

in some countries sqrt(-1) is wrong yeah

[livid horizon]

That's just i

[urban laurel]

cause sqrt(x) : [0, inf [ --> [0, inf[

[vast shale]

yes it's i

[livid horizon]

Ik

I was saying in general, for complex numbers too

But if the set is defined for real numbers only then it should be positive

[urban laurel]

$\int \frac{1 }{\sqrt{3z^2-4}}\ \dd z$

[twin meteor]

Adam Chebil

[urban laurel]

i want to factor out the 3 in denom

but z^2-4/3 can be negative

[vast shale]

Well all I know is that the property of splitting the square root only applies if both A and B are positive. And we all agree that in dealing with square roots of negative numbers, you need to work with imaginary numbers.

[hybrid flicker]

But z²-4/3 is negative if and only if 3z²-4 is negative

So assume both quantities are positive

[urban laurel]

oh that makes sense

thnx !

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[vast shale]

I'm trying to work on my little hobby programming project and need to implement an algorithm that sets a price value of a player depending on his age and rating.

I would like the rating to have the heaviest factor in the pricing, and the age has a lower weigth factor. Younger players with a high rating will thus be worth more than an old player with identical rating.

Is there anyone able to help me with this?

I can implement the algorithm myself programmatically, just need the algo first :p

[white whale]

What do you mean by price value?

I didn't understand the context of the problem.

What also is rating here.

You mean like Elo rating?

[empty frigate]

also you're being kind of vague and qualitative (just saying one factor is weighted "more"), you might need to like, play around with some different options to see which particular "shape" and which constants make the most sense with the kind of stats actual players have

a price assignment that seems fairly good might actually end up valuing age more in practice because the players are thousands of years old and you weren't expecting that
or depending on the rating system, maybe 3100 should actually be 200 times as valuable as 3000 because that's the separation between "fairly experienced" and the best in the world, or maybe it isn't worth much more at all because a difference of "100" is actually mostly just random noise from it being hard to measure how good someone really is

[vast shale]

The are caps on the age and rating limits.

There*

[empty frigate]

that kind of thing changes what you should be doing at a quantitative level, just knowing that "higher rating is better" doesn't distinguish between "if you're 200 points better you almost always win", "we only really care about differences of 10,000", "the value becomes infinite at exactly 4905 because those are the people who can play theoretically perfectly", "actually it's % differences that are important, not the absolute numbers; 600 to 700 is way bigger than 2000 to 2100" etc.

[vast shale]

I am not good at maths at all, I just figured sharing how I expected one to have a more important role in the price determination was important to mention.

The price in this case would be the currency paid to buy a player, such as in football contracts in real life.

A lower aged player can develop itself more meaning it has more potential and can be priced higher.

Etc.

[white whale]

So, it's a program to predict a footballer's contract value?

[vast shale]

A fictional footballer's contract value, yes.

[empty frigate]

do you have data on the contract values of other fictional footballers? since extrapolating from existing numerical data might be easier than tuning it by hand

[vast shale]

I'm generating everything from scratch

Generating the players "randomly"

And now need to assign a price value based on the generated rating and age.

 { country: 'Germany', rating: 75, position: 'Attacker', age: 37 }

Basically the data I'm working with

[empty frigate]

...ok in that case i think basically the issue is just that we need more information

[white whale]

It's a football game?

[vast shale]

Yes

[empty frigate]

like i could say that this player is worth $1,000,000,000,075,037

[vast shale]

I want to make an offline football manager game as a web app

localhost webapp 😄

[empty frigate]

each player is worth 1 quadrillion dollars, plus one thousand dollars * rating, plus one dollar * age

and this does technically do what you originally said, because rating matters more

[vast shale]

You're right. I want the prices to be capped at (25 000 - 75 000 000).

[white whale]

Can a player increase the ratings over time?

[vast shale]

I also suppose players don't rise in price linearly, that would make buying them too easy .

And yes, players (young ones faster than old) will develop during the game.

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[keen oyster]

Find the total surface area of the given combined solid:

[vocal sleet]

Please don't occupy multiple help channels.

[keen oyster]

help

.close

[vocal sleet]

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[vast shale]

Prove through mathematical induction :7|3^(2n+1)+2^(n+2), n∈ℕ

[white whale]

I mean, do you really need to? The question can easily be answered with basic modular arithmetics.

[vast shale]

The problem says by induction

.close

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[tardy python]

I need help 5x^2 +3x-2<0

[half imp]

!status

[vocal sleet]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@tardy python Has your question been resolved?

[vocal sleet]

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[hardy crater]

,rotate

[hardy crater]

Hello, I have stuck on part B

,rotate

[hardy crater]

I know which line I need to work out

As it also the height

Which I need for part C

To work out the area

I drew it on the thingy if you can see

[broken mesa]

The easy way to do it is to use the point-to-line formula

[hardy crater]

So if I work out the Equation of line PQ?

[livid horizon]

😁 🐻

[broken mesa]

you ever saw this equation?

[lime basin]

yes, then use the formula given above

[hardy crater]

Never seen before

Perpendicular is the opposite reciprocal

So could I do it that way?

[lime basin]

you can do it without that equation too, sure

[hardy crater]

And use the coordinates of Q

[broken mesa]

you can do it without the equation though

up to you

[hardy crater]

Never seen that equation before

Looks useful for the future though

[lime basin]

get the slope of SR, then negative reciprocal of that is the slope of the point Q joined perpendicularly to SR (D), then form the equation QD, then find the common point of QD and SR, that will be D, then find QD

[hardy crater]

Got it

Thank you all

.close

[vocal sleet]

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[vast shale]

Prove through mathematical induction: 3^(𝑛+1)≥𝑛^4+n^2+1, n≥5

[vast shale]

so, how do you prove something by induction?

you need the base case, inductive hypothesis, and step

we start from the base case

[urban laurel]

prove that the inequality is true for n = 5 first

[vast shale]

I can't do the induction step

[urban laurel]

!show

[vocal sleet]

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

[urban laurel]

what's the hypothesis ?

[vast shale]

P(k+1)?

We weren't taught like this

So I don't really know

[urban laurel]

Induction hypothesis : $3^{n+1}\geq n^4+n^2+1 \$ Prove that : $3^{n+2}\geq (n+1)^4+(n+1)^2+1$

[twin meteor]

Adam Chebil

[vast shale]

I was trying to prove the hypothesis

With 3k^4+3k^2+3≥(k+1)^4+(k+1)^2+1

I failed

<@&286206848099549185>

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[crisp saddle]

You don't prove the hypothesis; you assume/use it to prove the $3^{n+2}\ge(n+1)^4 + (n+1) ^ 2 + 1$.

[twin meteor]

poypoyan

[urban laurel]

Induction hypothesis : $3^{n+1}\geq n^4+n^2+1 \$ Prove that : $3^{n+2}\geq n^4+4n^3+7n^2+6n+3$

[twin meteor]

Adam Chebil

[urban laurel]

we know : $3^{n+1}\geq n^4+n^2+1 \ \implies 3^{n+2}\geq 3n^4+3n^2+3$

[twin meteor]

Adam Chebil

[urban laurel]

now we just need to prove that : $\ 3n^4+3n^2+3\geq n^4+4n^3+7n^2+6n+3$ ; for $n\geq 5 \ \implies 2n^4-4n^3-4n^2-6n \geq 0$

[twin meteor]

Adam Chebil

[vast shale]

Thanks!

[urban laurel]

well it's not solved yet xd

[vast shale]

.close

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[vast shale]

in this question, help me if im correct
in the cyclic quadilateral DB=BO+DO
DB = radius + radius
BO=radius and DO= radius
simmilarly: AO=radius and CO=radius
now since AOB is an icoselus triangle OBA=50 degree and AOB=80 degree
DOC=80 degree
now by SAS triangle AOB=triangle DOC
DCA=50 degree
now: since triangle OBC is also an isocelus tringle
OBC=60 degree = OCB
BCD=angle DCA+BCA
BCD=50degree+60degree
BCD=110

[verbal tiger]

How to solve this?

[vast shale]

um, i think

you should post this question in some other place

[sturdy pumice]

Channel is taken , use another one @verbal tiger

[verbal tiger]

Yeah, noob mistake

[vast shale]

still help needed

[sturdy pumice]

Its mentioned nowhere in the question that O is the center

[vast shale]

yea right

ok i blundered

tysm

.close

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[hazy hound]

When I find a choice for $N$ in an epsilon-delta definition proof, how or where do I fit in all that algebra showing $n \ge N \implies |x - L| < \epsilon$

[twin meteor]

o.O

[hazy hound]

ex. I have $\left{1 + \frac{1}{\sqrt{n + 2}}\right} \rightarrow 1$

[twin meteor]

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

[median crane]

\FT{
Let (\eps > 0) be arbitrary. Choose (N) such that blah blah.
Then,

[ \abs{x_n - L} \ldots ]
}

[twin meteor]

Pure

[left talon]

i like the blah blah 😄

[median crane]

[hazy hound]

I hav it setup like this

do I just take all that algebra and replace n with N

[median crane]

\FT{
Let (\eps > 0) be arbitrary. Choose (N) such that [insert your magical (N)]
Then for (n\ge N),
\begin{align*}
\abs{x_n - L} &\le \text{ here you manipulate the inequalities} \
& \phantom{w}\vdots \
& < \eps
\end{align*}
}

[twin meteor]

Pure

[median crane]

You can't just start off by saying (\wrb{\abs{x_n - 1} < \eps}) because that's what you want to prove.

[twin meteor]

Pure

[hazy hound]

so is like this satisfactory

[median crane]

That'd work, yes.

\FT{
Let (\eps > 0) be arbitrary. Choose (N) such that (N\eps^2 > 1)
Then for (n\ge N),
\begin{align*}
\abs{x_n - L} &= \abs{\df 1{\s{n+2}}}\
&< \df 1{\s n}\
&< \df 1{\s{\df 1{\eps^2}}}\
&=\eps
\end{align*}
}

This also works.

one sec i copied of tex so its bad

[twin meteor]

Pure

[hazy hound]

hmm yes I forgot I can make my life easier by doing that

[median crane]

[hazy hound]

wait why did you write Ne^2 > 1

[median crane]

What's wrong with it?

[hazy hound]

I was just curious why you didnt isolate N

[median crane]

Ah I see.

I just prefer to write it that way because it alligns better with my statement of the archimedean principle.

\begin{thm}[Archimedean Property of \(\R\)]
If \(x\in\R,y\in\R\) with \(x>0\), then there is a positive integer \(n \in \N\) such that \[ nx > y\] 
\end{thm}
\begin{proof}
Suppose, for the sake of contradiction, that there is no such \(n\). Consider the set \[ A \coloneq \left\{\; nx \; \middle|\; n \in\N \;\right\} \]
This set is bounded above by \(y\), and so there exists a least upper bound denoted \(\alpha \coloneq \sup A\). By definition of supremum, \(\alpha - x\) is not an upper bound for \(A\). Thus,
\[
\alpha - x < Mx \qquad \text{for some }M \in\N 
\] 
However, it then follows that \(\alpha < (M+1)x\), which is a contradiction since \(M+1\in\N\).
\end{proof}

[twin meteor]

Pure

[median crane]

Thats the theorem i have in ym notes

[hazy hound]

I only learned archimedian property of N and it was like just to say something is 0 and i never thought more about it

[median crane]

I mean it's the same thing. I just find this a bit more intuitive.

[hazy hound]

okay, thank you very much

[median crane]

You can't fully justify

such choice of N without ther archimedean property of R.

[hazy hound]

I see, thank you. I looked back at our textbook and I see something about this now

but I guess I don't see them explicitly saying they're using it to justify a limit proof

[median crane]

[hazy hound]

I have to go to Costco now. Thanks for all the help

.close

[vocal sleet]

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[median crane]

Okay. Good luck with that.

[vocal sleet]

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[rough depot]

I did the first step of taking deravitives

I get fx = y+z, fy = x+z, and fz = y+x

And the equation then would be something like:

[scarlet tapir]

use the formula

[rough depot]

(y+z)(x-1)+(x+z)(y-2)+(y+x)(z-5)=0

I'm blanking. Which formula?

[scarlet tapir]

fx(x0,y0)(x - x0) + fy(x0,y0)(y - y0) - (z - z0)=0

Put the values of the point (1,2,5) in this fx = y+z, fy = x+z, and fz = y+x

and then use the above formula

[rough depot]

I see. Ty ty 😊

[scarlet tapir]

np

[vale ermine]

Would replacing xy directly with (xy1 + yx1)/2 same as for yz and zx work?

[rough depot]

Using the method, I got 7x+6y+3z = 34

.close

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[charred shell]

Does anyone know how I can do this?

[charred shell]

Please ping me if anyone knows

It would help me a lot

<@&286206848099549185> thanks so much

[gilded spade]

Out of straight guess, if we assuming both values for x and y to be equal to 22 then we can assume that 1/x + 1y = 1/11 is technically correct

[charred shell]

yeah, but i need to find the number of solutions

[gilded spade]

I haven't done the other way through, found this through straight guess

[gilded spade]

Got it.

[charred shell]

I cannot thank you enough

this question is bugging me ever since i started it

[gilded spade]

No it's fine, it's just a little trick

[charred shell]

oh

[gilded spade]

At least i think so

hold on just a sec

[charred shell]

sure

thanks again

[tidal dock]

well technically there are infinitely many solutions for x and y

[gilded spade]

Exactly, but i dont think that's the answer in literal sense.

[charred shell]

so 1/11 can be written as a sum of 2 unit fractions infinite times?

[tidal dock]

but for $\lfloor x \rfloor$ and $\lfloor y \rfloor$ there are only finitely many

[twin meteor]

artemetra

[charred shell]

yeah

because x and y need to be integers

[tidal dock]

huh?

then floor does nothing

if x and y are integers

it's equivalent to solving $\frac{1}{x}+\frac{1}{y} = \frac{1}{11}$

[twin meteor]

artemetra

[gilded spade]

So if we sum up 1/x + 1/y, we may get y+x/xy, and then we put it as

y+x/xy = 1/11

Then

xy = 11y + 11x

x = 11y+11x/y or y = 11y+11x/x

I could be very wrong at this, so please pardon me

[tidal dock]

with a computer i just found $\frac{1}{-110} + \frac{1}{10}$ to work

[twin meteor]

artemetra

[charred shell]

but i need to find the number of solutions

[tidal dock]

i know

in general, what you are looking at is a diophantine equation

[charred shell]

yep

[tidal dock]

elaborating on what icarus wrote, $xy=11(y+x)$

[twin meteor]

artemetra

[tidal dock]

since 11 is prime, either x or y must have a factor of 11

[charred shell]

true

[tidal dock]

WLOG, let's say y=11k for some integer k

$11xk = 11(11k+x)$

[twin meteor]

artemetra

[tidal dock]

$xk=11k+x$

[twin meteor]

artemetra

[tidal dock]

$(k-1)x=11k$

[twin meteor]

artemetra

[charred shell]

ok

got it

soo

what's next?

[vocal sleet]

@charred shell Has your question been resolved?

[charred shell]

welp

not here, but i figured it out

so sure ig

[vocal sleet]

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[quaint creek]

hi

[quaint creek]

i am not sure what to do here

idk how i can apply the division algorithm

because i have to know that 0 <= r < |a|

but how would i know that 2 < |2L + 1|

[proven spade]

so you've shown that (2l+1,2l+3)=(2l+1,2)

[quaint creek]

well idk if i have because im not sure i correctly applied the algorithm

[proven spade]

can you use division algorithm again

[proven spade]

this step is correct

[quaint creek]

isnt it required for 0 <= r < |a| though

r = 2
a = 2L + 1

[proven spade]

yes

[quaint creek]

how do i know that 2 < 2L + 1

[proven spade]

but if else then you're done trivially since a=1

so the gcd must be 1

[quaint creek]

a is 1?

am i allowed to just switch around the values and take a to be 1 and q to be (2L + 1)

[proven spade]

no I'm saying if 2>2L+1 then a=1

[proven spade]

are you referring to the a in the question or the statement of the d.a (i.e. a=bq+r)

[quaint creek]

the a in b = aq + r

[proven spade]

do you know gcd(x,y)=gcd(x,y-kx)

[quaint creek]

im not sure if we were given that

the lemma it mentions in the question is gcd(a, b) = gcd(r, a)

[proven spade]

yes and r=b-aq

[proven spade]

so you can apply it twice

right?

since (a,a+2)=(a,2)=(1,2)=1 right

and we can do this since 2<a (if a=1 then it is obvious) , and 1<2

[vocal sleet]

@quaint creek Has your question been resolved?

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[bleak mango]

i dont understand this

dont you have to apply the product rule?

oh wait

never

mnd

.close

[bold hound]

they are doing that

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[near vigil]

how do i solve for x? how do i turn the equation with either sinx or cosx? i tried to rewrite sinx as cos(pi/2-x)

[left talon]

ye u can do that

[near vigil]

ye, but then what

i am stuck, i have no idea how to proceed after that

[left talon]

u agree with me that instead of 1 u can write:
cosx^2 + sin^2 ?

[near vigil]

yeah

[near vigil]

i also tried this

but doesnt help

[left talon]

k use the fact that cos^2 = 1-sin^2

do this:
3cosx = 1 - 2sinx

u agree?

now square both sides u will get:
9cos^2 = (a-2sin)^2 agree?

now use the fact that cos^2 = 1-sin^2

so u will get:
9 * (1 - sin^2) = (1-2sin)^2

keep going and in the end u will get:
13 (sinx)^2 -4sinx -8 = 0

assume that t = sinx

and solve for t

u will get:
13t^2 -4t -8 = 0

[near vigil]

you mixed up the numbers but i think i understood

[left talon]

find t then find x

[near vigil]

lemmy try

[left talon]

if you need anything else u can ask

[near vigil]

thank you, i solved it using your method, i mean i got the qaudratic equation

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[cerulean cove]

Can someone help me with this question

[paper depot]

what is troubling you?

[vast shale]

Volume of a cylinder = B*h

[paper depot]

gonna tap the sign preemptively

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[cerulean cove]

and would it be mm^3

[vast shale]

What’s mm?

[cerulean cove]

i meant m

[vast shale]

Yea

[cerulean cove]

As in unit

[vast shale]

Volume is ^3

[cerulean cove]

I did 17^2 X 14 and I got 4046 and It said it was wrong

[vast shale]

Why are you squaring 17?

[cerulean cove]

Am I not supposed to do that

[vast shale]

No

[cerulean cove]

Thanks got it

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[merry hemlock]

I can’t find my mistake😭 It’s PQ formula btw and my result here is wrong. The end result is supposed to be -8 and 7 but idk how I got that

[hoary blaze]

ur supposed to multiply the -14 too

if you're gonna *4

[merry hemlock]

Oh oops

Thank you!

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[vagrant solar]

A piece of bread is thrown vertically downwards from a cliff at 3ms-1. It is caught by a swooping seagull, 20m above the base of the cliff, 4s after being thrown. What is the height of the cliff

[vagrant solar]

If I do S=ut+1/2at^2 I get 90.4

but if I do v^2/u^2=2as I get 0.459

(gravity is 9.8 not 9.81 in this context)

Why is there a difference in the answers

s = x + 20
u = 3
v = 0
a = 9.8
t = 4

[dusty ice]

Why have you said v = 0?

[vagrant solar]

because the seagull stops the falling animation

[dusty ice]

to apply suvat equations, you need to have constant acceleration / no external forces

the seagull catching it rapidly decelerates the bread, as opposed to the previously constant acceleration due to gravity

[vagrant solar]

oh I see

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[rustic stirrup]

What did I do wrong? I mean I was not supposed to use cosecant because we haven't studied that yet but even without using it and going for the 1/sin^2(x) formula I still get -2cotg(2x)

[tribal moss]

hint:

$\int_{}^{}\frac{dx}{sin^{2}x\cdot cos^{2}x}=\int_{}^{}\frac{sin^{2}x+cos^{2}x}{sin^{2}x\cdot cos^{2}x}dx=_\cdots }$

[twin meteor]

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

[rustic stirrup]

🤔 Yeah that would work but isn't mine correct too?

[tribal moss]

your instructor wanted the most popular answer

tan x - cotan x

so eh cud negate your trial

[honest relic]

Nothing's wrong, unless I don't remember the identities correctly.
-2cot(2x) = -2(cos(2x)/sin(2x)) = -2( (cos²x - sin²x)/2sinxcosx) = -(cos²x/sinxcosx + sin²x/sinxcosx) = -cosx/sinx + sinx/cosx = -cot(x) + tan(x) .

[rustic stirrup]

Wait so it's the same thing then?

[tribal moss]

$\int_{}^{}\frac{dx}{sin^{2}x\cdot cos^{2}x}=\int_{}^{}\frac{sin^{2}x+cos^{2}x}{sin^{2}x\cdot cos^{2}x}dx=\int_{}^{}\frac{dx}{cos^{2}x}+\int_{}^{}\frac{dx}{sin^{2}x}=\\=tgx-ctgx+C$

[twin meteor]

Joanna Angel

[rustic stirrup]

Yeah I got that thanks but is mine incorrect then?

Cuz I've been trying to figure it out for 3 hours

[tribal moss]

yes 🙂 it seems all ok at oru calculations

but believe me

he or she wanted the classical answer

and he or she did not even

read

this part ofd yours work

in is or her opinion you did it improperly

[rustic stirrup]

Oh nobody's checking it I just wrote that over with red lol, i have the answers it's just for practice

He's usually checking answers very carefully I just wasn't sure if they're both correct

So -2cotg(2x) is also correct?

[tribal moss]

as an exercise, please transform yoru answer to mine

ok ?

start from your anseer and prove they are thr sme = an identiy

[rustic stirrup]

The milk dude kinda did that, but alright I'll try

[tribal moss]

great 🙂

[tribal moss]

$-2ctg2x = -2\frac{cos2x}{sin2x}=-2\frac{cos^{2}x-sin^{2}x}{2sinxcosx}=\\=-\left( \frac{cosx}{sinx} -\frac{sinx}{cosx}\right)=-\left( ctgx-tgx \right)=tgx-ctgx$

[twin meteor]

Joanna Angel

[rustic stirrup]

Kinda a mess but I got it, thanks

@honest relic ty as well

[tribal moss]

🙂

[rustic stirrup]

❤️

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[hollow rock]

hello need help with calculus problem

[lucid bane]

note the derivative describes the slope of the tangent line of at every point. Treat the derivative as a separate function and try to maximize it.

[hollow rock]

y' = 120x^2 - 15x^4

y"= 240x - 60x^3

is this right..

[vocal sleet]

@hollow rock Has your question been resolved?

[pallid forge]

yes

[hollow rock]

i dont know what to do next

[pallid forge]

well you want the max value of y' right? won't this occur when y''=0?

[hollow rock]

so x^2 = 4

?

x = 0, + or - 2..

[pallid forge]

yup

[hollow rock]

f ' 0 = 0

[pallid forge]

but which one is it

[hollow rock]

um

[pallid forge]

so the max occurs at x=0, x=2, and/or x=-2

[hollow rock]

yeah

[pallid forge]

the max of y'

[hollow rock]

240..?

im lost

[pallid forge]

yea!

that's it

that's the slope of the tangent line at it's max value

[hollow rock]

wait

so the slope, i mean the answet is just 240?

and what should I write down "what is this slope?"

[vocal sleet]

@hollow rock Has your question been resolved?

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@hollow rock Has your question been resolved?

[onyx flax]

Nice work! You can probably state "The maximal slope of 240 is attained when x=2 or -2"

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[devout spire]

My book says that when regarding if C ⊆ B, there is "insufficient data". The only information I get is this Venn diagram, nothing more.
My problem is that I would say that C is not a subset of B. Although they share the element "w", I'd say that C isn't a subset of B since in this Venn diagram (to me), it infers that there are more elements in B, that just weren't shown in the image.
My question is should I assume that in the set B, there could or could not be additional elements?

[dim stream]

i think the issue is that you're only given 2 elements, $t \in A$, and $w \in B, C$

[twin meteor]

marieme

[dim stream]

so, since there's no more information, we don't know if there is an element in C that isnt in B

if we only had these two data points, t and w, then C would be a subset of B

but a new data point could easily change this fact, so we need more info

[devout spire]

but the way set B isn't inside C, I'd think that one couldn't be a subset of the other.

i understand what you mean, but the drawing got me a bit confused

[dim stream]

that's simply due to the way the visual venn diagram is set up. it's misleading

ignoring the visuals and only looking at the hard data of the system, we can see that there's no way to say that C is or isnt a subset of B without more information

[devout spire]

there's no way to know because i don't have a "let C = {whatever}?
ahh, so it can be misleading. that's what i needed to know.

[ancient knoll]

the way ABC are set up is that B\C is currently empty

[ancient knoll]

!occupied

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

[dim stream]

yeah this

[devout spire]

hmmm

they are currently empty?

as in later they could change, or that really that all the information wasn't given?

well. i think i get it. thanks a lot

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[tulip tapir]

How do I do this? I tried parameterizing as r(u, v) = <cos(u), sin(u), 0>, but got that ru x rv was 0.

[tulip tapir]

I know I need to take the surface integral of curl, but I don't know the way to do it, or what surface to use.

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@tulip tapir Has your question been resolved?

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@tulip tapir Has your question been resolved?

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[vocal sleet]

Please don't occupy multiple help channels.

[left haven]

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[celest pilot]

True or False: If a linear program is infeasible, then its dual linear program is unbounded

[celest pilot]

I want to know what others think

Ok, so the answers say it's false, but I think that is a typo?

[late ingot]

it is possible for both the dual and the primal to be infeasible. See dual linear program for details and several more examples.

from wikipedia on Linear Programming. so i suppose it might be right because there are cases where the dual doesn't have to be unbounded with an infeasible primal? i haven't studied this so this is the best i can do. hopefully someone else can chime in soon.

[celest pilot]

I thought it may have been because the fact we know is "for any dual that's unbounded, its primal is infeasible".
And this is claiming the reverse, which I'm personally not familiar with

[late ingot]

i think its probing the fact for whether an infeasible LP guarantees an unbounded dual.

[celest pilot]

No idea tbh ;_;

[late ingot]

im curious, what course is this for?

[celest pilot]

Linear programming

[late ingot]

comp sci?

i wanna suggest that you post this in #optimization, they might have more people on your level.

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[eternal magnet]

Where does a_0 /2 come from in the final solution? Alls I see about Fourier is that it would lead to a_0

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<@&286206848099549185>

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[undone aurora]

so i know that

theta = (pi + 2pi)/6 * k

now do i let k = 0, 1, 2, 3, 4, 5
coz we need 6 angles?

like does it matter what 6 numbers I pick for k here?

[elder scaffold]

yo

pi + 2kpi/6

[undone aurora]

ye

thats what I got too

but then I have to pick 6 angles right

[elder scaffold]

it doesnt matter what values you pick for k as long as they are in succession

[undone aurora]

oh

alr ty

[elder scaffold]

though be careful when u pick like that

[undone aurora]

it has to be below 2pi right?

[elder scaffold]

depends what principal domain u are using

some people do [0,2pi] some use (-pi,pi]

[undone aurora]

oh alr

ty

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[gleaming eagle]

why would the leading coefficient of f9x)=1/3(x-4)(x+5)^2(x+1) be 1/3 ?

[gleaming eagle]

the biggest degree is on the (x+5)

... as I say that, I understand dsnaskd

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[elder scaffold]

because if you expand the brackets you will find that all the x's multiply out to x^4 and that will be our largest term, then that gets multiplied by 1/3 so our first term becomes 1/3 * x^4 , therefore the leading coefficient is 1/3

oh

wait u said u understand

[gleaming eagle]

... o h

well I thought I did

I thought it was cause it was in factored form o.O

oopsie

ty D

:D*

[elder scaffold]

all good

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[hasty pulsar]

General solution is y(x)= C1x +c2
For y(0)=0 we get 0=c2 and for y'(1)=0 we get C1 = 0
How come the particular solutions are y1(x)= x and y2(x)= 1?

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[hasty pulsar]

<@&286206848099549185>

[split token]

Yeah

[hasty pulsar]

i think i got it

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[chrome thorn]

Triangles ABC and ACD lie in different planes, where the line BD is perpendicular to the plane ABC, angle ACD is 90°, CB = 6√3 cm, and DC = 12 cm. Find the dihedral angle formed by the faces containing these triangles.

[vocal sleet]

Please don't occupy multiple help channels.

[chrome thorn]

90?

<@&286206848099549185>

[true totem]

What do we have to solve in this question @chrome thorn

please tell me i will try to help you

[chrome thorn]

Find the dihedral angle whose faces contain the given triangles

[chrome thorn]

Triangles ABC and ACD lie in different planes, where the line BD is perpendicular to the plane ABC, angle ACD is 90°, CB = 6√3 cm, and DC = 12 cm. Find the dihedral angle formed by the faces containing these triangles.

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[quick rain]

<@&286206848099549185>

[quick rain]

could someone help me finish this matrix. i’m having a tough time with it.

[livid horizon]

The result is a number right?

[quick rain]

yes

this is the original matrix form

[median crane]

[ \nmat[v]{a&b\c&d\} = ad - bc]

[livid horizon]

Wait how did you go from that matrix to finding determinants

bc*

[median crane]

Oof

[quick rain]

it says find the value of Z in problem 5 by determinants

[twin meteor]

Pure

[livid horizon]

Check?

[median crane]

Autocorrect

[livid horizon]

Lol

[livid horizon]

Oh

[quick rain]

and problem 5s linear combination was

2x+3y-z=-1
-x+5y+3z=-10
3x-y-6z=5

[quick rain]

so what do i do in the one up above?

[livid horizon]

Ohh

[quick rain]

^

[livid horizon]

And you've to find x,y,z?

[quick rain]

just z

[median crane]

NZ

[vast shale]

are you finding the determinant? do u know how to find the determinant of 2x2

[quick rain]

yes, been awhile.

[vast shale]

oh PR posted

[quick rain]

came here to get a refresh

[median crane]

!xy

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[vast shale]

do you understand this @quick rain

at least

[quick rain]

yes