help-49

[rain wasp]

are you in an exam?

[molten socket]

no

[rain wasp]

sus

[molten socket]

fair

[wet bridge]

can i say e^(cosh) approaches to e slower than the denominator

therefore 0/0 but denominator overweights

so 0

[molten socket]

yes the answer is indeed zero

[wet bridge]

my reasoning might be faulty (probably)

[ember marsh]

i have a solution without derivative

but i used taylor series...

[molten socket]

pls solution without derivative

bro ok

[wet bridge]

lol

[molten socket]

im just looking to cancel h in the denominator

[wet bridge]

wont happen

[molten socket]

nah fr why

or whatever limit techniquse i can use

[wet bridge]

to cancel it out you need a h in numerator

[molten socket]

e^x-1 over x or smth

[wet bridge]

l’hopital is a limit techique that would be perfect to use here

but its derivatives so

[ember marsh]

maybe not so rigorous

but

for small angles

$\cos h \approx 1 - \frac{h^2}{2}$

[grand pond]

BrandenXia

[ember marsh]

so we have

$e^{\cos h} \approx e^{1 - \frac{h^2}{2}} = e^{1} \cdot e^{-\frac{h^2}{2}} = e \cdot e^{-\frac{h^2}{2}}$

[grand pond]

BrandenXia

[wet bridge]

with approximation its obvious the limit is 0 anyways

[gaunt nimbus]

Approximation from mclaurin😭😭

Dude’s on limit

[flat moat]

At this point just use L'hopital, anyway these small angle derived from derivative

[ember marsh]

he don't know derivative🤣

[wet bridge]

why cant we find a solution

was it gonna be impossible if we didnt invent l’hopital

[gaunt nimbus]

It’s the limit definition of derivative 😭

[gaunt nimbus]

No

[wet bridge]

so a solution with only using the limits is possible

[ember marsh]

do we have (cosx-1)/x approachs 0 as x goes to 0

[wet bridge]

but the counterargument is derivatives are also only using limits

[molten socket]

why does lhopital work anyways so weird

[ember marsh]

emm

[molten socket]

my syllabus doesnt even teach lhopital coz the smartest students always manage to find the limit without lhopital

cringe ass

[ember marsh]

do u know the limit of sinx/x as x goes to 0?

[sullen jungle]

it only works with indeterminate forms and compares how much "faster" something is approaching 0 or inf

for example yeah sinx/x

[wet bridge]

so is this considered a solution

[molten socket]

yes

anyways it appears no solution from purely limits

[sullen jungle]

https://www.instagram.com/reel/DA8U_0cxwpy/?igsh=MW9yNHNyYmRyOXQxdg== funny proof of sinx/x without hopital

c.carre__

Haaland teaches Zlatan Sin x/x = 1 🔥 #zlatan #zlatanibrahimovic #ibrahimovic #haaland #math #calculus #engineering

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[gaunt nimbus]

Have u guys gotten the answer

[ember marsh]

with approximation yes, but not a rigorous one

[gaunt nimbus]

No limit definition?

[ember marsh]

one from intuition

[ember marsh]

one using approximation

[gaunt nimbus]

why not limit definition of derivative?

[ember marsh]

because he don't know derivative

[gaunt nimbus]

ahhh

[ember marsh]

🥲

it really depends on how we rigorously define sine function

[wet bridge]

?

[ember marsh]

if we just define it with series, the solution will be obvious

[flat moat]

Can we use squeeze theorem tho?

[gaunt nimbus]

no bound

[flat moat]

Wait

nvm

[ember marsh]

emm

maybe it will work

if we consider left and right limit seperately with squeeze theorem

something like this?

[gaunt nimbus]

how?

[sullen jungle]

yea

you can do thatp

[midnight plank]

@molten socket Has your question been resolved?

[midnight plank]

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[jovial crest]

hi! can someone explain what injective and surjective functions are? ( not a definition but in a manner that could explain how we use them in solutions )

[rain wasp]

can you show a problem?

[jovial crest]

yeah sure

i did solve it but i found another solution which uses surjectivity and couldn't understand

[rain wasp]

what does the question ask?

[jovial crest]

find f(x)

[humble torrent]

Simply put, if you say had the equation f(x) = y, then if

• f was injective, then there’s at most 1 solution for x

• f was surjective, then there’s atleast 1 solution for x

[jovial crest]

in this solution could someone tell me why f is surjective? and f(0) was assumed to be 0, cause its injective ( if someone could explain how that works too that'd be wonderful )

[jaunty canopy]

a function is injective if it's 1-to-1 basically

[humble torrent]

Saying f(0) = 0 because f is injective sounds very misleading btw

[jaunty canopy]

i.e. if i gave you some real number y, there is at most one real number x such that f(x) = y

[jovial crest]

yeah i understood the definitions but not how they were used in this solution

[jaunty canopy]

ah k

i see ur doing functional eqns lol

[jovial crest]

oh

[jovial crest]

yup, only really know number theory and combinatorics so trynna learn this so i dont miserably flunk the national mo

[jaunty canopy]

so they've plugged in (x, -f(x)) to get f(0) = 2x + f(f(-f(x)) - x)

so that means f is surjective

because f(0) is some constant

the image of f must be R because of ur exposed 2x

(i.e. alternatively you can think of it as if i gave you what f is)

[jovial crest]

huhh

[jaunty canopy]

(you can find f(0) - 2x = f(f(-f(x)) - x))

(so if ur dedicated enough, you can invert that to get x = f(something) )

(so f is surjective)

[jovial crest]

how

[jaunty canopy]

they haven't used f(0) = 0 anywhere yet, and also injectivity doesn't imply f(0) = 0

so as an example, let's sps f(0) = 0 cus it makes our calculations nicer

we know f(0) - 2x = f(f(-f(x)) - x)

i.e. -2x = f(f(-f(x)) - x)

can you tell me a value of z such that f(z) = 3?

can u tell me a value of z such that f(z) = pi?

[jovial crest]

i think?

[jaunty canopy]

as in can u tell me?

[jovial crest]

im sorry i dont follow

[jaunty canopy]

or maybe f(z) = -4

[jovial crest]

can u explain it to me as though i have no idea what surjectivity or injectivity is besides the basic definition

[jaunty canopy]

so -2x = f(f(-f(x)) - x)

[jovial crest]

like from the top explain why f is surjective

[jaunty canopy]

can u tell me a value of z where f(z) = -4?

[jovial crest]

im so sorry but ive been clueless this message down

[jaunty canopy]

so right, we plug in (x, -f(x)) into our original functional equation

that gets us f(0) = 2x + f(f(-f(x)) - x)

[jovial crest]

yes

[jaunty canopy]

from here, you can immediately conclude f is surjective if you were writing this up in a contest

but let's go through why it's surjective step by step

[jovial crest]

yes pls tysm

[jaunty canopy]

for the moment, we'll suppose we've also shown f(0) = 0

this isn't necessarily, but it makes my maths slightly nicer

so we actually have 0 = 2x + f(f(-f(x)) - x)

[jovial crest]

wait how are we assuming this

[jaunty canopy]

as in this is just as an example

[jovial crest]

ohh

[jaunty canopy]

of why it's surjective

[jovial crest]

alright

[jaunty canopy]

i haven't shown that, i'm just gonna assume it because it makes algebra simplify a bit nicer

[jaunty canopy]

so we rearrange this to get

-2x = f(f(-f(x)) - x)

following?

[jovial crest]

understood so we're just assuming it as 0 for the sake of brevity and itll give us the same conclusion regardless?

[jovial crest]

yes

[jaunty canopy]

so do you remember what the definition of surjectivity is?

[jovial crest]

yes if f is surjective there's atleat 1 soln

[jaunty canopy]

yeah so we want to show that

for each $y \in \mathbb{R}$, $\exists z \in \mathbb{R}$ such that $f(z) = y$

[grand pond]

LY

[jovial crest]

yes

[jaunty canopy]

we'll do it for a few specific examples of y

how you do it generally should be clear

[jaunty canopy]

so let's say y=-4

i want to find some real number such that f(z) = -4

what happens if we plug x=2?

[jovial crest]

-4=f(f(-f(2))-2)

[jaunty canopy]

so can u tell me a value of z such that f(z) = -4?

[jovial crest]

huh

[jaunty canopy]

yeah so f(-f(2)) - 2 is a solution to f(z) = -4

[jovial crest]

ohh right

[jaunty canopy]

can u tell me a solution to f(z) = -6 now?

[jovial crest]

-6=f(f(-f(3)-3)
so f(-f(3))-3 ?

[jaunty canopy]

yep

anyway yeah so that's why f is surjective

[jaunty canopy]

solving this directly will be a bit annoying

[jovial crest]

because there is a solution?

[jaunty canopy]

but it'll be like f(0) - 2x = y

x = -(y+f(0))/2

so f(-f(x)- x) will be a solution to f(z) = y where x = etc.

[jaunty canopy]

yh

it doesn't matter that we don't know what real number it is, surjectivity just says that some number exists

and we've shown that the solution exists

[jovial crest]

so then what would injectivity say?

[jaunty canopy]

as in for injectivity, you need to show that if f(x) = f(y), then x=y

anyway the first functional equations you encounter are probably "just plug stuff in until it works"

injectivity and surjectivity are normally the 2 things you encounter next and are quite useful

for surjectivity, often we can assume i.e. "let y be such that f(y) = 0" or something like that

[jaunty canopy]

in general, if your functional equation has exposed variables (i.e. see the exposed x out in the open), that's normally how you could prove injectivity and surjectivity

on the other hand, if everything in your functional equation was wrapped up in an f, although you might still be able to prove injectivity/surjectivity, it'll be a lot harder

[jovial crest]

OHHHH I UNDERSTAND

[jaunty canopy]

so for problems like this where there is an exposed x, you want to ask yourself "can we prove injectivity and/or surjectivity"

[jovial crest]

so a function can be injective and surjective simultaneously?

[jaunty canopy]

yh

an injective and surjective function is a bijective function

a bijection is like one of those normal functions where you can "invert" it

(i.e. f inverse exists)

[jovial crest]

so f(x)=k has a solution, and f(x)=f(y) iff x=y?

[jaunty canopy]

yh

[jovial crest]

okk so i have this one question which involves injectivity and surjectivity can you help me with that

like just one question so i know how to use it when solving questions

[jaunty canopy]

sure

[jaunty canopy]

anyway now that we've shown f is surjective, can you show f is injective?

we have an exposed x, and we want to wrap all xs in an f all by itself

[jaunty canopy]

anyway this sort of motivation will come up a lot in olympiads

as an example, here, everything is wrapped in an f, so it'll be hard to prove injectivity/surjectivity (and ||in fact, the only solutions are constant so you wouldn't be able to prove injectivity/surjectivity||)

[jovial crest]

i tried but im not getting anything of value how would i do that

[jaunty canopy]

on the other hand, for this problem, if you assume f is not constant, then you can prove f is injective but it's quite hard to do since there is no exposed x/y

[jaunty canopy]

so the functional equation is f(f(x) + y) = 2x + f(f(y) - x)

for injectivity, it always goes "suppose f(a) = f(b)"

like i mentioned, the exposed x will be useful in trying to prove f is injective

so we should try x=a and x=b

that get's us f(f(a) + y) = 2a + f(f(y) - a)

and f(f(b) + y) = 2b + f(f(y) - b)

wait nvm my idea doesn't work

let me try it properly

[jovial crest]

okk

[jovial crest]

im not getting anywhere either so if u get anywhere help me out

<@&286206848099549185>

[jaunty canopy]

btw i have to dip soon but if ur still here later i can help u with ur query

[jovial crest]

ill probably check back here in some time tag me when you're here

[jaunty canopy]

what is the original problem btw/where is it from?

[jovial crest]

imo 2002 shortlist

[jaunty canopy]

oh huh it's only A1

[jovial crest]

ye

[jaunty canopy]

ok i'll look into it later, gtg now but another good problem for injectivity/surjectivity if i remember correctly is Balkan MO 2000 P1

[jovial crest]

will do, thanks!

[jovial crest]

can i just dm you the other queston i need help with ( also funcitonal eqns )? since this help channel will get occupied soon

[midnight plank]

@jovial crest Has your question been resolved?

[midnight plank]

@jovial crest Has your question been resolved?

[jovial crest]

<@&286206848099549185>

[midnight plank]

@jovial crest Has your question been resolved?

[midnight plank]

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[midnight plank]

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[frank osprey]

Can anyone tell me why we can ignore things which are multiplied by X if X approaches 0? I mean its not literally 0 so why can we completely ignore it and still have an equality

[frank osprey]

Like a perfect equality, not just a really close approximation

[prime hornet]

you mean $\lim_{x \to 0} cf(x) = c \cdot \lim_{x \to 0} f(x)$?

[grand pond]

higher!

[frank osprey]

Do i? Idk

Like just if i have a 5x + c and x approaches 0, then its equal to c

[prime hornet]

do you have an example to describe what you mean?

[frank osprey]

Here

[exotic pelican]

do you know the basic idea behind limits?

[frank osprey]

I still did not learn about limits

[sharp tulip]

Wouldn't it be because the as x -> 0, 5x -> 0

[frank osprey]

Yeah but 5x approaches 0, but why is it perfectly equal to 0?

If X is not perfectly equal to 0

[sharp tulip]

Because it's a limit

[exotic pelican]

that's a definition thing

we define the lim to be what the inner term approaches

[frank osprey]

How can i define that a + b = b if a is small

[sharp tulip]

Limits find the value that a function or expression approaches, not what it actually equals

[frank osprey]

I mean i just dont understand why can we say its perfectly equal not just a very close approximation

[exotic pelican]

we don't define 5x + c = c, we define lim(5x + c) = c

[frank osprey]

Ohhh

[sharp tulip]

yea

[frank osprey]

Ohhh ok now i understand

Thank you guys!!!

[sharp tulip]

I get you, math is weird

[frank osprey]

Lol

Ty ty :))

[sharp tulip]

Bye

[frank osprey]

Cya

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[midnight plank]

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[midnight plank]

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[last slate]

ok so here's my reasoning

$$T_\text{total} = T_{R \to M} + T_{M \to R}$$

[grand pond]

Edmund Cloudsley

[last slate]

hence

$$T_\text{total} = \frac{2.4 \times 10^{12}}{c} + \frac{2.4 \times 10^{12} - (0.90c) \times \frac{2.4 \times 10^{12}}{c}}{c}$$

[grand pond]

Edmund Cloudsley

[last slate]

The rocket would have moved the distance of 0.90c times (2.4 \times 10^12) over c

in the time span that the light hits the mirror

am I correct or my reasoning flawed?

I am fairly certain that my value for $T_{R \to M}$

[grand pond]

Edmund Cloudsley

[last slate]

is correct

however I am not so sure about T m to r

rough sketch I made of the situation

apologies for the bad handwriting

<@&286206848099549185>

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[last slate]

.close

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[safe onyx]

why aren't these the same

[safe onyx]

(windows calculator)

[radiant roost]

The integral isn't unique, there's a family of solutions

It's showing you one of them

[safe onyx]

oh so it choses to make the constant 1

🤔

desmos choses to make it 0

¯_(ツ)_/¯

[radiant roost]

It seems so

[safe onyx]

thank you :D

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[midnight plank]

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[radiant roost]

I think this statement is a little imprecise

[midnight plank]

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[main stone]

Hello!

[main stone]

136 students were asked how many presents they got on christmas, there were given following ranges and votes by each.
What's the average present amount and median?
0 - 16 votes
1 - 4 votes
2-3 - 18 votes
4-5 - 26 votes
6-7 - 20 votes
8 or more - 52 votes

[next rover]

no idea

"more" messes up everything

[midnight plank]

@main stone Has your question been resolved?

[safe narwhal]

0-1 - 20 votes
8-9 - 52 votes
?

[main stone]

...

dont troll

[nimble copper]

they aren't trolling

those 52 students could have each received 8 gifts. Those same 52 students could have each received 52 gifts

[main stone]

would it be accurate to not count the 52 votes?

its

,calc 5700/136

[grand pond]

Result:

41.911764705882

[main stone]

42% of votes

[radiant roost]

i think you can still find the median

not the mean though

[nimble copper]

indeed the median is unaffected

[main stone]

why is median not unaffected?

[nimble copper]

begone double negative

what is your working definition of median?

[main stone]

ok how do i calulate median?

in this

cus i know how to calculate it with normal numbers

1,1,1,2,3

mediam is 1

[radiant roost]

i'd probably find the total number of votes first

[main stone]

136

[radiant roost]

then try to split it in half

[main stone]

,calc 136/2

[grand pond]

Result:

68

[radiant roost]

actually you still can't find the median

because it says 2-3 presents

you don't know if it's 2 or 3

sorry i meant 6 or 7

it would fall in the 6-7 range i think

[main stone]

so midpoinr?

[radiant roost]

ye

the midpoint

[main stone]

Ok do that for me

<@&286206848099549185>

[midnight plank]

@main stone Has your question been resolved?

[midnight plank]

@main stone Has your question been resolved?

[spiral osprey]

We’re not supposed to do work for you.

[main stone]

Or?

[spiral osprey]

!nosols

[midnight plank]

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.

[spiral osprey]

Or more appropriately

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[midnight plank]

@main stone Has your question been resolved?

[hot narwhal]

How can I help?

[midnight plank]

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[wet bridge]

AB=ED, BC=BD

[dreamy lichen]

there is an isosceles triangle hidden

you could fill in most angles using that one

i dont yet know how it will be helpful (bc i havent solved it yet) but its probably a good way to get started

[wet bridge]

its not going anywhere from that

i tried it

[twilit jetty]

did you figure out angle DEC and angle EDC?

[wet bridge]

yes

those are the all known angles

[dreamy lichen]

You could probably just use trig

but thats long and tedious

[wet bridge]

and i want a geometrical solution

[dreamy lichen]

solution is ||25 degrees|| by the trig shit, but idk how to get there. Maybe the answer will help

[eager zephyr]

What are you exactly trying to find in the figure?

[wet bridge]

angle x

[wet bridge]

the drawing is also to scale

[eager zephyr]

bro take sum of all angles of a quadrilateral = 360

[twilit jetty]

that doesnt work

you can already see in each triangle that it adds to 180

so doing that does not give any new information, you get 180 + 180 = 360 as expected with nothing to solve for x

if AB = ED was not given, there would be infinitely many solutions for x (0 < x < 120)

[eager zephyr]

what are the wavy symbols on bc and bd supposed to mean?

[twilit jetty]

those indicate that BC = BD, theyre supposed to stand in for tickmarks

[eager zephyr]

oh ok

[wet bridge]

i can say x=25 if i can prove E’DBA is a cyclic quadrilateral

probably not a good way

[eager zephyr]

what is the total value of angle E?

[wet bridge]

95

bed is 95

[eager zephyr]

it is not a cyclic quadrilateral then.

opposite angles don’t add upto 180

[wet bridge]

ae’d is supposed to be 120 if thats the case

but i cant prove

it is the case according to the answer

[eager zephyr]

is there a total area given?

[wet bridge]

no

[wet bridge]

<@&286206848099549185>

[drifting root]

what are we tryna do?

fimd x?

[wet bridge]

yes

[twilit jetty]

using this picture, and given that BC = BD, you have to figure out what x is
trig says x is ||25 degrees|| but there should be a purely geometric way to solve this

[wet bridge]

yes

im also looking for a geometrical solution

anyone?

[midnight plank]

@wet bridge Has your question been resolved?

[wet bridge]

<@&286206848099549185>

[hot narwhal]

Use the angle sum rules for triangles and quadrilaterals

[wet bridge]

well guess what

tldr that doesnt work

[wet bridge]

.

[tawdry laurel]

Meanwhile the diagram:

[wet bridge]

the diagram is TO THE SCALE

[tawdry laurel]

The negative angle;

[hot narwhal]

Aight

[wet bridge]

also i know x=25

i dont want the result i want the solution

no trig

wdym how

[wet bridge]

refer to this

[midnight plank]

@wet bridge Has your question been resolved?

[wet bridge]

<@&286206848099549185>

[midnight plank]

@wet bridge Has your question been resolved?

[wet bridge]

<@&286206848099549185>

[warm elm]

Qn?

[ebon anchor]

hi

how cna i help you?

[bleak berry]

this gives x = 25 if you can prove if F E C are all coliniear @wet bridge

[bleak berry]

since FBD is equilateral

[wet bridge]

colinear because FBC is an isosceles triangle

niceeeeeeee

x=25

wait lemme put it in steps

[bleak berry]

is just that enough to show its colinear though

like what if it looked like

[wet bridge]

this is enough

select a point F such that BAF are colinear and AF=BE

since BF=BD and <FBD=60, FBD is an equilateral triangle

now construct FE

<CFB+<BCF+<FBC=180 so F, E and C are colinear

and finally triangle BAD and FED are congruent

therefore <CFD=<ADB=x=25

[wet bridge]

also just solve for x

[bleak berry]

you dont know that its 35 though

[wet bridge]

BF=BC also

also

i cant draw it lol

draw the circle with center B and radius BD

it’ll pass through F and C

CBD=50 central angle

so CFD=25 so BFC=35

[crude galleon]

are you done?

[wet bridge]

.close

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[main tusk]

Given the stepfunction:
how can I find y(t) here?

[main tusk]

Someone made this, but I dont know how to set this up /and if this is even correct

[static terrace]

Define the step function so we're on the same page if possible

Also our names match 🥰

[main tusk]

Wym define the step function

Im just given the graph, thats all.

[static terrace]

You using the conventional u(t) or?

[main tusk]

yes

u(t) being the heaviside

[static terrace]

Right so the first part of our graph needs to be a linear function

With coefficient 3 right?

[main tusk]

Yes

[static terrace]

So 3t, but we need to make sure it only shows up for t from 0 to 1

[main tusk]

*u(t)

[static terrace]

If we multiply it by u(t) we ensure it starts at 0

[main tusk]

and for cutting it off, doing - then

[static terrace]

Yes

[main tusk]

okay okay

[static terrace]

With an offset of 1

Apply same logic for all other segments

[main tusk]

Is it normal I get this in desmos?

[static terrace]

Hmmm, note that the other segments need to be offset vertically since the don't start at t=0

[main tusk]

and how would you offset them vertically then?

[static terrace]

(2(t-1) +3)

Try this

For the second segment

[main tusk]

(2(t-1) -3) seems to work

hold on

Looks more like it

[static terrace]

Eyup

[main tusk]

So this is not correct then

[static terrace]

Now simplify your expression if you want and that's it

I guess not

[main tusk]

pretty complex to wrap my head around

im guessing doing it segment per segment is the way to go

[static terrace]

No time to verify, sorry, but your graph looks correct

Yeah, that's how you should always do it

[main tusk]

Okay, thanks!

(nice name btw)

[midnight plank]

@main tusk Has your question been resolved?

[midnight plank]

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[drifting root]

this is what i have so far, idk how to go on

[viral dagger]

cant you split them?

[drifting root]

if i split them, what do i do with the sec

oh

do i do the stupid sec(secx+tanx)/(secx+tanx)?

[sullen jungle]

probably something like that i forgot it

,w integrate secantx

[grand pond]

Results provided by WolframAlpha

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

Requested by _rysel_

[drifting root]

hmmm

ok

i think i got it

thx

.close

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[viral dagger]

determine all positive real $x$ such that for all positive reals $a,b,c$ atleast one of $a+\frac{x}{ab}$, $b+\frac{x}{bc}$, $c+\frac{x}{ca}$ is greater than or equal to $x+1$

[grand pond]

skissue.in.a.teacup

[viral dagger]

is this am-gm or smth

[visual tiger]

I'm guessing you could try to prove their sum is greater than 3x+3

[viral dagger]

(a+b+c)+x(a+b+c)/abc

this doesent seem useful

[pine thicket]

this is a trick question i think

you can always find a counterexample

hint: think of simple cases

[viral dagger]

like 1,1,1?

[pine thicket]

indeed

[viral dagger]

wait uhh

its geq

[pine thicket]

darn

[viral dagger]

oh boy do i hate myself some incorrectly written question (this is my 10th streak)

[visual tiger]

yeahh

ok I found x <= -4

I don't know how

I don't know if it works for every x <= -4

but I know it's a necessary condition

[viral dagger]

x is a positive real :(

[visual tiger]

oh

wait

ah I see where I messed up

not this one

I forgot the solution x = 1/2

ok x = 1/2 seems to be correct

my hint:

[viral dagger]

wait whats stopping us from just choosing b=x+2

[visual tiger]

if it's true for all a,b,c>0 then it's true for this particular value of b

but what are you going to do with it

[viral dagger]

oh right

[visual tiger]

anyways

my hint is

if you pick a = b = c

then you must have 3(a+ x/a^2) >= x+1

[visual tiger]

also just a quick disclaimer I now realize I forgot the 3 so we might need changing that

[visual tiger]

if this is true for all a > 0, it's true for the input where the function f(a) = a+x/a^2 attains its minimum on (0,infinity)

-> find a = a(x) the argmin

[visual tiger]

solve the inequation that you get plugging a(x)

[viral dagger]

whats argmin

[visual tiger]

f(argmin) = min

the input(s) that produces the min

arg-min for minimum argument

[viral dagger]

the minimun f(a) is at 1-2x/a^3=0=>a=cbrt(2x)

if we plug back we get 3(cbrt(2x)+x/cbrt(4x^2))>=x+1

[visual tiger]

sorry this is more adequate

[viral dagger]

what am i looking at

[visual tiger]

the range of values of a

[viral dagger]

how did it become complex

[visual tiger]

the result is not complex

when you simplify

[viral dagger]

wtf

what is the range supposed to represent

[visual tiger]

well the a(x) for which this is true

gives us this in the end

not very pretty

how can we even check if those values work for all a,b,c

oh wait

no I was right beforehand

x = 1/2 is the only answer because

I thought I missed a three

but it cancels

[viral dagger]

ok not gonna lie how do you get that

[visual tiger]

(a+ x/a^2) >= x+1

[viral dagger]

i still have no idea

[visual tiger]

I wrote 3(a+ x/a^2) >= x+1

but the x+1 add up too

(a+ x/a^2) >= x+1

[viral dagger]

oh that makes more sense

[visual tiger]

after dividing by 3 on both sides

so

with a = cbrt(2x)

we get a^3 - 3a + 2 <= 0

notice the nice factorization incoming

[viral dagger]

(a-1)(a-1)(a+2)

[visual tiger]

so

(a-1)^2 * (a+2) <= 0

if a > 0, not a lot of solutions

[viral dagger]

wait

isnt it like tangent at a=1

for a>0 its only a=1 no?

[visual tiger]

well yeah

a+2 > 0

so (a-1)^2 <= 0

so a-1 = 0

a = 1

-> x = 1/2

[viral dagger]

thats the only solution?

[visual tiger]

the only possible one, yeah

[visual tiger]

we know any x solution must verify this for all a > 0

so (cbrt(2x) + x/cbrt(2x)^2) >= x+1

so 3/2 * cbrt(2x) >= x + 1

[viral dagger]

oh its equality

[visual tiger]

etc...

[visual tiger]

we don't know that

[viral dagger]

what if a,b,c arent equal?

[visual tiger]

again, we've only shown x = 1/2 is the only POSSIBLE solution

is it a solution

idk

we haven't seen what happens for a,b,c different

[viral dagger]

what but at the start you made the presumption a=b=c, if proving x=1/2 works for a=b=c that doesent mean its the only solution as a whole no?

[visual tiger]

jic we're not on the same page

possible solution doesn't mean solution

it means "can be a solution"

[viral dagger]

oh whoops sorry

[visual tiger]

what we've shown so far

x is a solution => x = 1/2

I never proved

x = 1/2 => x is a solution

wait

[viral dagger]

1.2+(1/2)/(1.2)(1.4)<1.5

id this a counterexample

[visual tiger]

well no

cause you have 1.4 + (1/2)/((1.2)(1.2)) >= 1.5

most likely

[visual tiger]

you only need at least one to be greater or equal to 1.5

[viral dagger]

oh right

[visual tiger]

also the justification for this I just need to alter slightly

just plug in a = b = c to those quantities

once of which is greater or equal to x+1

all of those quantities are a + x/a^2

anyways back to now

take x = 1/2

[visual tiger]

do we have one of those bigger or equal to 3/2 no matter a,b,c>0

if we managed to prove (a+b+c)(1+ 1/(2abc)) >= 9/2 we'd be done

unfortunately this is really tricky

[viral dagger]

wait

[viral dagger]

this no?

[visual tiger]

how did you get that

[viral dagger]

am gm

a/2+a/2+b/2+b/2+c/2+c/2+x/ab+x/ac+x/bc

[visual tiger]

mmmh

so this is bigger or equal to 9 * root3(x/4)

bigger or equal to 9/2

well done

equality case when a = b = c = 1/(2a^2), meaning a = b = c = cbrt(2)

[viral dagger]

so done?

[visual tiger]

I will head to bed now

[viral dagger]

alr ty! gn

.close

[midnight plank]

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[dawn dagger]

good night

[midnight plank]

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[thorn sorrel]

is ctg cotangent?

ive never seen that notation

[ionic magnet]

yes

ctg = contangent

[cerulean oyster]

my way to solve it would be to translate the problem into tan 3x >= 1

[delicate sage]

cot^-1 (1)

[ionic magnet]

why you changed sign

[delicate sage]

= pi/4

[cerulean oyster]

because if tangent is greater than 1, then cotangent is smaller than 1

[lean sequoia]

for me is pi/6

aaaa

no

no

im wrong

sorry

[cerulean oyster]

given that tangent is =1 when cosine and sine are the same, youll have to check both points in which they take the same absolute values.

But since youre looking for greater than, you can discard the negative portion

[lean sequoia]

tg 3x>=tg pi/4

[cerulean oyster]

usually, try to not give the answer straight away, but explain what you did, or hint to it

[lean sequoia]

okkkk

[cerulean oyster]

Youll find that you essentially want to maximize sine in relation to cosine, since tan = sin/cos.

if you look at the regular graph on tangent, is really easy to find how, basically take values increasing from pi/4 up to pi/2, it will later repeat for each tangent repetition.

youll have to consider that youre working with 3x, not x, also.

[midnight plank]

@ionic magnet Has your question been resolved?

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[pastel tree]

$f(x,y) = \sqrt{x^2+y^2}$

[grand pond]

Merineth 🇸🇪

[pastel tree]

If i'm asked to sketch the graph of this given R^3 function

How would i go about it?

I understand that it's equivalently saying:

$z = \sqrt{x^2+y^2}$

[grand pond]

Merineth 🇸🇪

[pastel tree]

Could this be solved the same way in R^2? Where we input values into x and y to determine z ?

[thorn sorrel]

wouldnt thag be 3 dimensional

[pastel tree]

Yes, hence the R^3

@tribal temple Are you sleeping? :>

I do realize that at z = 0 the point is at the origin (0,0,0)

[twilit field]

yes

start by squaring both sides

[pastel tree]

yeah i realize that it becomes z^2 = x^2 + y^2

I'm just not entirely sure how to sketch it

I realize that the blue point represents where it starts

[twilit field]

Well, it's a set of circles, isn't it?

[pastel tree]

but for example if x = 1 and y = 1 then z = sqrt(2)

[radiant roost]

consider cross-sections with constant z

[burnt flame]

these are called level curves :)

[pastel tree]

Not entirely sure i follow that cross section part

do you mean we set z to 1, 2.. etc?

[radiant roost]

yeah

[pastel tree]

x^2 + y^2 = 1
x^2 + y^2 = 4
...

[burnt flame]

it also helps to set x or y equal to 0. this gives you a sense of what the function looks like in the yz or xz plane

[pastel tree]

wouldn't this be drawn in 2d?

[robust isle]

yeah but then you superpose all the level curves, that gives you the function you want

[twilit field]

well, draw the shadow of the curve, so to speak

[pastel tree]

Hm this is a bit complicated

[robust isle]

well you need to believe in your vision skills yeah

[tribal temple]

(I was )

[burnt flame]

it is better to think in cross sections with these sorts of things. i think the combination of setting x=0 and y=0, and the set of level curves will help paint a picture of what the surface is :)

[robust isle]

🎶 and then I saw all her faces, now I'm a believer

[pastel tree]

Hmm okay

so how does the cross section method work for sketching?

$x^2+y^2 = 1$

[grand pond]

Merineth 🇸🇪

[pastel tree]

solve for x and y respectively?

[burnt flame]

no

in the original function, set x or y equal to 0

[pastel tree]

ok

$z^2 = x^2$

[grand pond]

Merineth 🇸🇪

[burnt flame]

technically if you were using this equation, it would be z=sqrt(x^2)

are you familiar with the absolute value function?

[pastel tree]

z = +-x ?

[burnt flame]

if x is positive, yes

but what happens if x is negative

[burnt flame]

@pastel tree

[pastel tree]

not much

Ok i had no idea

$z = \sqrt{x^2} \implies z = |x|$

[grand pond]

Merineth 🇸🇪

[pastel tree]

I assumed it was

$z = \pm x$

[grand pond]

Merineth 🇸🇪

[pastel tree]

or even

z = x

square both sides gives z^2 = x^2

[burnt flame]

if it were z^2=x^2 yes

but squaring adds extra solutions sometimes, so its a bit dangerous to do

[pastel tree]

$z = \sqrt{x^2} \implies z^2 = x^2 \implies \pm z = \pm x$

[grand pond]

Merineth 🇸🇪

[burnt flame]

take x=1, if we square both sides we get x^2=1, adding the solution of x=-1

[pastel tree]

Is this allowed?

[burnt flame]

does this make sense?

[pastel tree]

no

[burnt flame]

all the solutions of z=sqrt(x^2) will be within z^2=x^2, but not all the solutions of z^2=x^2 will be valid for z=sqrt(x^2)

[pastel tree]

I mainly want to learn to sketch functions

[burnt flame]

for example z cant be negative, as sqrt() only outputs nonnegative numbers

do you know how to sketch the absolute value function?

[pastel tree]

no

[burnt flame]

|x| is basically just the magnitude of x, it gets rid of the negative on negative numbers

|3|=3, |-3|=3, |-5|=5, |0|=0 (zero has no magnitude)

so on positive side, it would just look like x

[pastel tree]

I get this part

i just don't understand how this applies to the sketch

if z = 1

then the radius in the x and y axis would be 1 for the circle?

[burnt flame]

when you set y=0, you get what the function looks like in the xz plane

[pastel tree]

but since we are setting z to a fixed value we get what it would look like in the x y plane?

[burnt flame]

it would just be a point in the xy plane if z=0. the only (x,y) pairs that satisfies it is (0,0)

[pastel tree]

ok i see what i'm doing wrong

i'm trying to draw it in x,y,z plane

but it's actually an x,y plane

[burnt flame]

the blue is the yz plane (where x=0), the red is the xy plane (where z=0), the purple is the xz plane (where y=0)

[pastel tree]

Yeah that is completely wrong according to the solution

[radiant roost]

the circles will go in planes parallel to the red plane

[pastel tree]

this is apparantly the solution

[radiant roost]

ok it looks like they just drew the level curves, they didn't draw a 3d picture

[pastel tree]

Ye seems so

[burnt flame]

what

[pastel tree]

very unclear exercise from the book

[burnt flame]

this isn't even what you said 😭

[pastel tree]

ohwell i'm just glad its solved

thanks

.close

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[gusty falcon]

how can i use the average rate of change here?

can i not?

[grand pond]

mmmm7

[fervent ember]

thats the average change for 25 years.

[gusty falcon]

yeah so what?

[fervent ember]

tkae the 5th root.

[gusty falcon]

😭 i'm confused

can you elaborate

wdym 5th root

where did that come from

[fervent ember]

you need an average for a five-year span.

you have to break down the value for 25 years to a 5 year period.

[gusty falcon]

right correct

didn’t notice that

[gusty falcon]

still don’t get this though

[fervent ember]

how would you combine a rate of cange for a first year with a rate of change for the second year to a two-year rate of change?

[gusty falcon]

not sure actually

i want to say compute the average of those 2 but that’s wrong

[fervent ember]

anyway. calculate the AROC 1935 -> 1940, then 1940 -> 1945, and then take the average f this 5 values.

[gusty falcon]

Okay no this one i know

[gusty falcon]

16-> 20 = 25%
20-> 30 = 50%
30-> 50 = 66.67 %
50->80 = 60%
80->150 = 87.5%

i know you can average all these

is there a way to only do (f(b) - f(a))/(b - a)

like for example (87.5-25)/(1960 -1935)

this doesn’t work but what’s the counterpart that works

[fervent ember]

this. AROC ^(1/5).

[gusty falcon]

can we do a related question first; maybe my question isn’t that clear

[fervent ember]

ask.

[gusty falcon]

here the answer is just (f(1980) - f(1972))/ (1980 - 1972)

right?

[fervent ember]

yes

[gusty falcon]

can i make a similar “graph”

for this question

and use AROC

that’s my question

[fervent ember]

in the tv question there is an average percentage asked, thats te difference to the student question.

thats why - to be honest - the use of AROC isnt correct here

[gusty falcon]

yeah well the y value is clear for the tv question

16-> 20 = 25%
20-> 30 = 50%
30-> 50 = 66.67 %
50->80 = 60%
80->150 = 87.5%

not sure about the x value

[gusty falcon]

hmmm but a percentage is just a decimal

looks like exactly the same question to me

the first x value will map to 0.25 for example

second x value will map to 0.5

and then you have a curve like the student problem

and then (0.875 - 0.25)/ (x_2 - x_1) would be AROC

no?

[fervent ember]

16->20: 16x(1+0.25)
20->30: 20x(1+0.5)
and so on.
so 16->150 is 16 x (1+0.25)x (1+0.5)x(1+066666)x....

[gusty falcon]

yes that’s what i wrote here

[fervent ember]

you do another calculation. a division, not an subtracton.

[gusty falcon]

the numbers i wrote means that much percent increase

[gusty falcon]

where?

[fervent ember]

20/16-1 = 0.25

[gusty falcon]

yeah but that’s to make a new table column which gives you

the percent increase

seemingly we want to use that column

to find the AROC

so yes additional calculation but so what? that’d for a new table column

[fervent ember]

sorry, you have to understand, that you do another calculation.
in the students example: number of students at the end minus the students at the beginning.
in the tv example: number of tv at the end divided by the number of tv at the beginnung.

[gusty falcon]

No i get that but 😭 we do that extra calculation to let’s say have another column labeled percent increase

imagine that column is for

[gusty falcon]

these values

[fervent ember]

and becaue of this:
you can divide by n (=number of years) to get the average.
and in the tv example you have to take the n-th root.

[fervent ember]

do it. what do you get?

[gusty falcon]

0
0.25
0.5
0.67
0.6
0.875

would be the column labeled as percent increase

[fervent ember]

0 is wrong.

you dont have a value at this position.

[gusty falcon]

hmm yeah

well eh

it’s 16-> 16 so one could argue there’s a 0% increase?

[fervent ember]

0.25 is a value from 1935 to 1940. so you would need a value for 1930 to calculate an change to a previous value.

you mix up values for a date/year with values for a period.

[gusty falcon]

okay so

i can’t use aroc because i have 5 inputs

and only 4 outputs?

so it isn’t a function

if it isn’t a function i can’t calculate the average rate of it

that?

[fervent ember]

you have 6 base values, so you could calulate 5 changes.

[gusty falcon]

well wait it’s a function but one input doesn’t map to any output

[gusty falcon]

so is my reasoning for why i can’t use the aroc thing right?

[fervent ember]

is said. do it. what do you get?

[gusty falcon]

i did

5 outputs 6 values

[fervent ember]

whats your AROC in the way you wanna calulate it?

[gusty falcon]

so i can’t do aroc: f(1960)- f(1935)/ (1960-1935)

but f(1935) isn’t defined

hence AROC isn’t defined

if u do it student way right?

[gusty falcon]

okay anyway the answer is just the average of these percents i guess

[fervent ember]

you could start with 1940 instead of 1935.

[gusty falcon]

then is (f(1965)-f(1940))/ (1965 - 1940) correct?

[fervent ember]

no, its not. but i wanna show you where you end up with your way.

[gusty falcon]

so yeah what’s the issue

[fervent ember]

what do you get if you calculate it this way?

[gusty falcon]

like 2.5%

[fervent ember]

do you think this result is realistic for a change frrom 15 to 160?

[gusty falcon]

yeah it’s not

but what’s causing the issue 😭

[fervent ember]

sorry to say, its this:

the AROC you use is for data series where you can add/subtract.

[gusty falcon]

😭ngl it doesn’t make much sense to me

i guess i can trust u on this but also because i don’t know what’s going on

[gusty falcon]

like is this supposed to be a fact?

[fervent ember]

yes. i said it before:

so 16->150 is 16 x (1+0.25)x (1+0.5)x(1+066666)x....

or in general: startvalue x (1+f1) x (1+f2) x (1+f3) x ...x (1+fn) = endvalue

if you wanna calculate the "average" of f1, ...fn you would calculate
average rate: (endvalue/startvalue)^(1/n) -1

its like geometric mean vs. arithmetic mean

in your student example you have value for 1972 + decrease in 1973 + decrease in 1974 .... = value in 1980
so you can calulate ( endvalue - startvalue) / years

[midnight plank]

@gusty falcon Has your question been resolved?

[gusty falcon]

Hmm thank you

i understand a bit of it

I don’t have a paper rn with me

Maybe i’ll have a look later and figure this out fully

thank you for now

[midnight plank]

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[fervent ember]

youre welcome

[midnight plank]

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[last slate]

can i do

(x^2-2x) ln e * (2x-2)= 0

??

[sharp coral]

no

[last slate]

oh

then how do i solve this?

[sharp coral]

if you have two things which multiply to 0 then one of them must be 0 individually

[last slate]

OHHH

HAHAHAHA

[dawn dagger]

zero product theorem

[last slate]

yea so x =1

but

lets say it wasnt 0

how would i solve it?

[dawn dagger]

then you would have a hard time to isolate x

algebraically, to say the least

[surreal moon]

You would be "very sad", i think is the mathematical term

[last slate]

thanks

.close

[midnight plank]

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[covert crypt]

YO

[covert crypt]

we had this example to find eigen values and eigen vectors.
when I tried to get eigen vectors I just can't get the same result

for example for lambda=2
I got [5/2k
k ]

I don't understand how we got 5k and 2k

specifically I am talking here

[woven dock]

What you have is equivalent. Since $a$ is arbitrary (i.e. can take on any real value) if you take $a\to2k$ you get the same as the whiteboard result

[grand pond]

Tymelord14

[covert crypt]

but what's the point of taking it as 2k?

[woven dock]

As far as I can tell, none other than aesthetics. It looks nicer to not have a fraction in the vector components

[tacit rose]

Fyi if you multiply by the matrix by 2 (to remove fractions) you'll see

[covert crypt]

I mean one of them should be equal to k and we should get the other in terms of k right?

[covert crypt]

oh

so we just did multiplied by 2 to look nicer

[tacit rose]

Basically, because [(5/2)k, 1k] is the same as [5k; 2k] since you just multiplied by a scalar

[covert crypt]

I got it now.

@woven dock @tacit rose Thank you for helping me

imma close this now

.close

[midnight plank]

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[viral dagger]

#help-32 message

[viral dagger]

a is done so only b

LY suggested to use fermsts christmas theorem so
(x-k/2)^2+(y-k/2)^2=k^2/2
(2x-k)^2+(2y-k)^2=2k^2
claim: if you can write 2k^2=a^2+b^2 where a≠b then k is beautiful
if k is odd, then mod 4 the rhs is 2 and the lhs is (even-odd)^2+(even-odd)^2=(odd)^2+(odd)^2=1+1=2
if k is even then mod 4 the rhs is 0 and the lhs is even^2+even^2=0 mod 4 so you can choose a,b that fulfills it, making it beautiful (proven)
if the only way to write 2k^2=k^2+k^2 then x=y which isnf possible so k is ugly
let p=1 mod 4, by christmas theorem
p = (x+yi)(x-yi)
2p^2 = (1+i)(1-i)(x+yi)^2(x-yi)^2
2p^2=(x^2-y^2+2xy)^2+(x^2-y^2-2xy)^2 so p is beautiful

any multiple of a beautiful number is beautiful

[raven locust]

is b done

i wanna do b

[viral dagger]

mg look at the second message :(

[raven locust]

its hard to read typing math without a clear font

[viral dagger]

for this, if (a,b,c) is a pythagorean triple, and x=(a+b+c)/2 and y=(b-a+c)/2 then you get (x^2+y^2)/(x+y)=c (there should be a 2 in the denom for the image)

oh wait

so the hypotenuse of a pythagorean triple is a beautiful number

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

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[viral dagger]

aa

[viral dagger]

#help-49 message

part b

LY suggested to use fermsts christmas theorem so
(x-k/2)^2+(y-k/2)^2=k^2/2
(2x-k)^2+(2y-k)^2=2k^2
claim: if you can write 2k^2=a^2+b^2 where a≠b then k is beautiful
if k is odd, then mod 4 the rhs is 2 and the lhs is (even-odd)^2+(even-odd)^2=(odd)^2+(odd)^2=1+1=2
if k is even then mod 4 the rhs is 0 and the lhs is even^2+even^2=0 mod 4 so you can choose a,b that fulfills it, making it beautiful (proven)
if the only way to write 2k^2=k^2+k^2 then x=y which isnf possible so k is ugly
let p=1 mod 4, by christmas theorem
p = (x+yi)(x-yi)
2p^2 = (1+i)(1-i)(x+yi)^2(x-yi)^2
2p^2=(x^2-y^2+2xy)^2+(x^2-y^2-2xy)^2 so p is beautiful

the hyp of a pythagorean trip is a beautiful number, not sure about the back part

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[jaunty canopy]

so we've proved that if p is a 1 mod 4 prime, p is beautiful

[viral dagger]

[jaunty canopy]

so that means all numbers divisible by 1 mod 4 primes are beautiful

what does that tell us about ugly numbers?

[viral dagger]

that its either divisible by 2 or is 3 mod 4?

[jaunty canopy]

yeah

so now we want to prove that if k is a product of 3 mod 4 primes and 2s

then k is ugly

i.e. suppose 2k^2 = a^2 + b^2

use fermat's christmas theorem to show a=b