help-39

[dusky scaffold]

Because it could only be one of these

[quick star]

no

it can be any of those?

hmm

[dusky scaffold]

Right so one of them is true

[quick star]

it can be any of those 4

[dusky scaffold]

yes

[quick star]

but she had

9 options to choose from

so it's 4/9

[dusky scaffold]

Not really

because those other options are neglected

that is

they arent "options"

because they arent possibly true

for example

how can the total be 2 balls

*beads

if jane has 5

the total has to be at least 6

jane would never choose 1, 2,3,4 or 5

@quick star makes sense?

[quick star]

i mean maybe

1/4 isn't the right answer regardless

[dusky scaffold]

really?

[quick star]

yes lol

[dusky scaffold]

Oh

Well

One thing is that John couldnt have had 1 bead

because if he did

the maximum possible total would be 6

he would never guess 8

similarly he would never guess 8 if he had 2 beads

so he must have had at least 3 beads

but he cant have had 3 beads since 8 was wrong

so he could only have 4 or 5 beads

so Jane can only guess 9 or 10

so 1/2

@quick star How about now?

[quick star]

hold on

yeah i think that makes sense

i can't check the answer yet

yeah if john guesssed 8 then he must've had at least a 3

but since 8 is wrong then 3 is wrong

so john has a 4 or a 5

[dusky scaffold]

https://tenor.com/view/manchester-united-cristiano-ronaldo-cr7-wallpaper-gif-25163312

[quick star]

and jane has a 5

so either 9 or 10

1/2

smart

i didn't see through all of that lol

i would lose at this game

[dusky scaffold]

https://tenor.com/view/nah-id-win-gojo-nah-id-win-gojo-satoru-gojo-jujutsu-kaisen-gif-722934852497975468

[quick star]

okay anyway

thank you!

.cllose

.close

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[ripe geyser]

can someone help me with question 3 and 4, the question set is about optimizations

[pearl pond]

@ripe geyser Has your question been resolved?

[ripe geyser]

<@&286206848099549185>

[pearl pond]

@ripe geyser Has your question been resolved?

[fluid axle]

try asking in #optimization @ripe geyser

[ripe geyser]

no access

[fluid axle]

go in id:customize and grab the undergrad role

this will get you access

@ripe geyser

[ripe geyser]

ty, so i asked in there, do they respond in my help?

[fluid axle]

well hopefully they do

not guaranteed obviously, but the guys who really know a particular topic don't go through the help channels usually

it's not "someone will respond", it's "there you'll have better luck having your post being seen by someone who understands what's going on"

@ripe geyser

[ripe geyser]

i see

[fluid axle]

if you describe what you've managed to do, and where you're stuck specifically, you'll also have a higher chance being answered

and if no one answers there, try math stackexchange or something

@ripe geyser

[ripe geyser]

ty for answering

[pearl pond]

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[sturdy sinew]

nvm

[sturdy sinew]

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[spring crystal]

What I do wrong. I didn’t know what to do after the u(1+u^2) times du/sec part

[sharp vigil]

i would suggest a different substitution

[spring crystal]

Wym

[sharp vigil]

set u = some other function instead of continuing with this one

[spring crystal]

So for these problems I’m supposed to just guess what i choose for u

[old marsh]

I think u = tan(theta) works fine tbh

[spring crystal]

So where did I go wrong

Also if I set u to sec there is no derivative of just sec?

[old marsh]

I would’ve just kept the integral as tan*sec^3

$\int \tan x \cdot \sec^2 x \cdot \sec x , dx$

[old marsh]

Then u = tan x

$\int u \cdot \sec x , du$

Following @spring crystal ?

[spring crystal]

Yes

[old marsh]

Can we write sec(x) in terms of u

?

[spring crystal]

Wym

[old marsh]

Because our integral is terms of ‘u’ now

if u = tan x

Then sec x = ?

[spring crystal]

Idk

[old marsh]

u = tan x

How can u isolate x

[spring crystal]

What the x is part of the tan tho

[old marsh]

Ok, but u can isolate it tho

There’s a special inverse function

Familiar with inverse trig?

[spring crystal]

No

[old marsh]

Ok then ur probably better off choosing a different sub

I’m shocked u haven’t learned inverse trig but ur in calc

[spring crystal]

Can u tell me

[old marsh]

well there’s no point if u won’t understand the next few steps, but yea I mean if

u = tan x
arctan (u) = x

Then sec(x) = sec(arctan(u))

Then we can simplify from there

But I don’t wanna confuse u so just choose a different approach

There’s another sub u can try which is u = sec(x)

[spring crystal]

Idk du of sec tho

[old marsh]

Du of sec is sec * tan

Yea both subs work perfectly fine

u = sec x is a lot shorter

u = tan x transforms into the other solution at the end quite nicely

[spring crystal]

I got tanx +c? Is that right

[old marsh]

No

[spring crystal]

I did u=sec^2

[old marsh]

That’s another great sub

But ur answer is still wrong

[spring crystal]

Actually I got (sec^2)^2/2

[old marsh]

That’s still wrong

u = tan(x)
u = sec(x)
u = sec^2 (x)

All good subs

[spring crystal]

What I do wrong

[old marsh]

du of sec^2 isn’t sec * tan

[spring crystal]

@old marsh is the answer sec^3/3

[old marsh]

Yes, but u don’t need me to verify that

Integral-calculator.com exists

[spring crystal]

@old marsh how about this

Idk what to do

[old marsh]

Why don’t u just go to the website I said

This one is also another algebraic manipulation one

[pearl pond]

@spring crystal Has your question been resolved?

[pearl pond]

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[tiny jay]

help

[tiny jay]

i cant do the first one'

i have this but idk what to do now

[pearl pond]

@tiny jay Has your question been resolved?

[pearl pond]

@tiny jay Has your question been resolved?

[midnight haven]

One thing

Is det(A+B) = det A + det B ?

Nvm

[pearl pond]

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[untold vine]

could someone explain how this works? i get the simplification of 2022C80+2022C81 into 2023C81 but after that? thanks

[hollow cobalt]

In general, nCr = nC(n-r)
Particularly, 2023C1943 = 2023C(2023-1943)

[untold vine]

oh yes hadnt spotted that - thank you :)

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[stoic imp]

let $f : \mathbb{R} \to \mathbb{R}$ differentiable such that $f(7) = 1$, $f'(7) = -2$ and $\\int_0^7 f(t) dt = 7$. Find the 2nd degree taylor polynomial of $x_0 = 1$ of the function: $$\ g(x) = \frac{x^3}{3} - \int_0^{x^2 + 6x} f(t) dt$$

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, P_2(x) = g(1) + (x-1)g'(1) + \frac{g''(1)}{2!}(x-1)^2

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, g(1) = \frac{1}{3} - \int_0^{1+6} f(t) dt

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, g'(x) = \frac{1}{3} \cdot \frac{d}{dx} \left[x^3 \right] - \frac{d}{dx} \int_0^{x^2+6x} f(t) dt \ g'(x) = \frac{1}{3} \cdot 3x^2 - f(x^2 + 6x) \cdot (2x+6) \ g'(x) = x^2 - (2x+6)f(x^2 + 6x) \ g'(1) = 1 - 8f(7)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

f(7) = 1

,align g''(x) &= \frac{d}{dx} \left[ x^2 - (2x+6)f(x^2 + 6x)\right] \ &= \frac{d}{dx} \left[x^2 \right] - \frac{d}{dx} \left[(2x+6)f(x^2 + 6x)\right] \ &= 2x - \left[(2x+6)'f(x^2 + 6x) + (2x+6)f'(x^2 + 6x) \cdot (2x + 6)\right] \ &= 2x - \left[2f(x^2 + 6x) + (2x+6)^2f'(x^2 + 6x) \right] \ &= 2x - 2f(x^2 + 6x) - (2x + 6)^2 f'(x^2 + 6x)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

g''(1) = 2 - 2f(7)-(64)f'(7)

g''(1) = 2 -2-64f'(7)

g''(1) = -64f'(7)

g''(1) = -64 x (-2) = 128

g'(1) = 1 -8 = -7

,, g(1) = \frac{1}{3} - \int_0^{7} f(t) dt \ g(1) = \frac{1}{3} - 7 = \frac{1}{3} - \frac{21}{3} = -\frac{20}{3}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, P_2(x) = g(1) + (x-1)g'(1) + \frac{g''(1)}{2!}(x-1)^2 \ P_2(x) = -\frac{20}{3} + (x-1) \cdot -7 + \frac{1}{2} \cdot 128 \cdot (x-1)^2 \ P_2(x) = -\frac{20}{3} -7(x-1) + 64(x-1)^2

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

.solved

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[heavy hill]

could someone help me figure out how to get 0 instead of -8 for this proof by induction

[plush bramble]

why are you doing induction in the first place

does the question specify to use induction

[heavy hill]

hey again!

yeah i have to use induction

[pearl pond]

@heavy hill Has your question been resolved?

[dusk narwhal]

Doing induction for this question is a bit weird but at that point one thing u could do (which is a bit dodgy) is aay that would be tru for n>=5

And then case study for n is 1 2 3 4

And then induct form that point

[heavy hill]

how would you prove that n >= 5

[dusk narwhal]

If n >=5 then 2k-10 >=0

So ur proof above would work for n >=5

Then u just do 1 2 3 4 manually

[heavy hill]

ohhh

so is there no way to prove it without having to do it this way

[dusk narwhal]

There probably is some obscure way

But i given the originaly difficulty of the question it just feels rly unnecessary

[heavy hill]

alrighties!

thank you so much for your help

.close

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[plush island]

Hi guys I’m looking for a mentor in maths , so I can become a « monster » at math and reach Olympiads . Anyone can help me?

[plush island]

<@&286206848099549185>

[lament cargo]

yups i can a little bit

[plush island]

Thank you

Can I dm you ?

@lament cargo

[lament cargo]

yups u can but i'll when i'll be free lol

i am a too busy tyoe of man

[plush island]

Yeh sure whenever you are free 😁

[lament cargo]

lmao

[pearl pond]

@plush island Has your question been resolved?

[dapper kraken]

this server is more oriented for individual questions unfortunately, so i reccomend maybe try going to a diffrent one or ask questions here about some questions

[pearl pond]

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[tender gate]

I am not very sure if the graph i made is correct or not

[pearl pond]

@tender gate Has your question been resolved?

[tender gate]

.close

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[real epoch]

https://tenor.com/view/rick-roll-gif-23663323

[pearl pond]

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[cobalt seal]

hello

[bronze coyote]

which question do u need help with

[cobalt seal]

the one with to the power a negetive sign

.close

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[rough bloom]

Hello !
Let’s take 4.5 x 10 to the power of -10 as an example

Since the exponent -10 is a negative whole number, you will need to move the decimal point of 4.5 ten times to the left. Don’t forget to plug in the zeros after you move the decimal point !

[rough bloom]

Here is how to do it 🙂

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[cinder flower]

this is so cute

[pearl pond]

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[rough forge]

"That was a good idea" shouts a sales man gladly. "I reduced the price for one pair of socks that was originally $1 per pair. In total, I made $109.61 selling these socks."$\$
"How many pairs of socks did you sell?", asks the chef.

[jolly parrot]

bacc (unhelpful)

[rough forge]

I made some progress

#help-20 message

and I would like someone to check it

The idea was to consider the amount in cents, then there exists a unique pair (p,n), and apply the fundamental theorem of arithmetics

integer pair

actually natural numbers

[oak quiver]

Why is the chef asking a sales person?

[rough forge]

In English

[pearl pond]

@rough forge Has your question been resolved?

[cursive wraith]

congrats you got the correct values

[rough forge]

thank you, it was due to your trick

howsoever you could tell to write it as a difference of squares

[cursive wraith]

oh yeah there are tricks to factorisation

otherwise you would have had to check every prime number below 100

and if you went in increasing order...

You would have had to check up until the very last one

however after checking "obvious" primes like 2 to 11

if you told yourself that maybe the cost wasn't lowered that much from 1 euro

you could have started checking from 97 and gone down from there

would have happened pretty quickly

[rough forge]

hmm ok

i just hope i am not screwed for the exam

i am not that good at puzzles haha

anyway thank you!

2003 gang

.solved

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[gentle pawn]

I don't get what exactly do they mean by "n" here and the things on the side are the options of the answer it's an mcq

[tall flint]

impossible to say without further information

what's the topic of this thing?

[gentle pawn]

numerical analysis

more accurately interpolation

[tall flint]

so then it would depend on precisely what interpolation method you're concerned with

[gentle pawn]

newton-gregory

i applied and am getting f(0.2) as 177.8 but idk even know what am i supposed to find

[tall flint]

how was newton-gregory defined? That's where you should look for what n is

[gentle pawn]

am sorry am not getting what do you mean?

[tall flint]

how was newton-gregory defined?

[gentle pawn]

f(x) = y0 + PΔ y0 + p(p-1)/2! + p(p-1)(p-2)/3!

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[somber summit]

guys how to do this

number 10

I haven’t started anything yet

[rigid lava]

if the side of the square is x cm

total amt off wire needed to make it will equal to ?

[somber summit]

50

total perimetre is 50

[rigid lava]

no no

[rigid lava]

in general

[somber summit]

What is amt

[rigid lava]

amount

[somber summit]

4x

It one side is x total amt is 4x?

Idkk

[rigid lava]

apply same logic to circle of radius "r"

[somber summit]

I got a and b

I don’t know how to do C

[pearl pond]

@somber summit Has your question been resolved?

[pearl pond]

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[midnight haven]

does F(x) in this question "Solve f(x) = x+ F(x) , f(2x) = x^2+2x find f(3x)" represent the integral of f(x)?

[cyan lily]

that would be my guess

[midnight haven]

Well

if not integration, then what?

[hard kite]

that's usually what it means

F' = f

[midnight haven]

yes but the thing is i don't know integration

neither does my friend

he gave me this question

Can't F(x) simply be another function?

[hard kite]

i mean, you could find f(3x) using just the second info given about f(2x)

[wet swallow]

It could be but it is most likely integral

Of f(x)

[midnight haven]

ok

.close

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[upbeat mirage]

The surface area of a cylinder is increasing at a rate of $9$
‍

square meters per hour.
The height of the cylinder is fixed at
‍

meters.
At a certain instant, the surface area is $36$
‍

square meters.
What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?

$\begin{aligned}
S(t)&=2\pi [r(t)]^2+2\pi r(t)h
\\
&=2\pi[r(t)]^2+6\pi r(t)
\end{aligned}$

[jolly parrot]

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

[upbeat mirage]

I don't get why you can relate S(t) like that

I think it should be S(r(t)) = that stuff

[eternal tulip]

its just notation

r is a function of t so S(r) is also a function of t

Ex: f(x)=x^2, x(t)=t^2 => f(t)=x(t)^2=t^4

[upbeat mirage]

But in this case

you can do f(x) = f(t^2) = t^2^2 = t^4

[eternal tulip]

yes

but t is the variable

[upbeat mirage]

right so

f(x(t)) = t^4

how does this mean

[eternal tulip]

you want your function to typically be written in terms of the root variable

like

[upbeat mirage]

f(t) = t^4

[eternal tulip]

this is just saying that f is a function of t, since x is also a function of t

well

[upbeat mirage]

no

it's saying that

f(t^2)=t^4

[eternal tulip]

yes

i see what you are saying

[upbeat mirage]

here it makes sense

[eternal tulip]

but you're not really looking at the pure function S(r) on its own

you're always looking at it in terms of S(r(t))

[upbeat mirage]

how

they said S(t) not S(r(t))

[eternal tulip]

they are redefining the function so its purely in terms of time

[upbeat mirage]

but the math should work out if you use S(r(t))

but if you take the derivative

you get S'(r(t))r'(t)

and since S'(t) is constant

S'(r(t)) should also be constant right

[eternal tulip]

take the derivative of their S(t) and you will see it is the same result as if you do d/dt(S(r(t)))

[upbeat mirage]

It doesn't work

with S(r(t))

S'(r(t))r'(t)=4pi(r(t))(r'(t)) + 6pir'(t)

the r'(t) cancel out

S'(r(t)) = 4pir(t) + 6pi

and then S'(t) = 9pi, so S(r(t)) = 9pi

and then you get r(t) is constant

[eternal tulip]

[S^{\prime}(t)=4\pi [r(t)]r^{\prime}(t)+2\pi r^{\prime}(t)h=r^{\prime}(t)(4\pi [r(t)]+2\pi h)]

[\f{d}{dt}S(r(t))=r^{\prime}(t)S^{\prime}(r(t))=r^{\prime}(t)(4\pi [r(t)]+2 \pi h)]

[upbeat mirage]

Right and in the second case

the r'(t) cancel out

[jolly parrot]

🫎Moosey is not Mοοsey🫎

[eternal tulip]

????

[upbeat mirage]

you have

r'(t)S'(r(t)) = r'(t)(stuff)

[eternal tulip]

yes

[upbeat mirage]

so you can cancel the r'(t)

S'(r(t)) = stuff

[eternal tulip]

no

S'(r(t))

not

[upbeat mirage]

yeah my bad

[eternal tulip]

S(r(t))

[upbeat mirage]

S'(r(t)) = stuff

So with this

since you know the derivative of S'(t) = 9pi

[eternal tulip]

no

[upbeat mirage]

the derivative of S'(r(t)) = 9pi

[eternal tulip]

well

[upbeat mirage]

sorry

S'(t) = 9pi

not the derivative of S'(t)

[eternal tulip]

9pi?

[upbeat mirage]

ye

[upbeat mirage]

here

it's fixed at 9pi

[eternal tulip]

says 9 here

[upbeat mirage]

ohh

I wrote it down wrong

[eternal tulip]

S'(t)=9

[upbeat mirage]

it should be 9pi

[eternal tulip]

ok

[upbeat mirage]

Let me send an image

[eternal tulip]

so you know S'(t)=9pi

[eternal tulip]

yes

[eternal tulip]

now, since you're asking about volume as well, you should try to write down a similar formula for volume of cylinder

[upbeat mirage]

Yeah, but before I do that

[upbeat mirage]

this implies that

r(t) is constant

which can't be true

[eternal tulip]

okay.

they should've wrote S(r)

to be less confusing

[upbeat mirage]

but it makes sense

saying that

the surface area at time t

[eternal tulip]

but they are meaning to say that Surface area is a function of time

[upbeat mirage]

is this thing in terms of radius at t

so S(t) and S(r(t)) are different functions

same function I guess

S(t) and S(r) are different

[eternal tulip]

yes

S(t) is S(r(t))

S(r)=2pir^2+2pirh

S'(r)=4pir+2pih

[eternal tulip]

which is why S'(r(t))=4pir(t)+2pih

make sense?

[upbeat mirage]

this makes sense

then S'(r(t)) = 9pi right

[eternal tulip]

no

S'(t)=9pi

[upbeat mirage]

why

not S'(r(t))

[pearl pond]

@upbeat mirage Has your question been resolved?

[pearl pond]

@upbeat mirage Has your question been resolved?

[upbeat mirage]

<@&286206848099549185>

[pearl pond]

@upbeat mirage Has your question been resolved?

[pearl pond]

@upbeat mirage Has your question been resolved?

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[rustic laurel]

Does my answer properly prove it?
It seems mostly right but the 2nd line from the bottom is congruent to x mod 6, is that an issue?

[pearl pond]

@rustic laurel Has your question been resolved?

[pearl pond]

@rustic laurel Has your question been resolved?

[ionic harness]

Yeah, thats not correct.

You have taken the wrong element. You could take s1=-x, s2=-x-1.

[pearl pond]

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[rustic laurel]

How do i know to take those?

[pearl pond]

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[sudden badger]

i need help with this identity

[lavish portal]

hmm, this looks like

you might want to use this formula

but the opposite of it 😭

[sudden badger]

for which part

[lavish portal]

the start?

use it for both parts, sin^4t and cos^4t

here is the one i was talking about

do you see how this formula comes from the cos(2x) double angle formula

[sudden badger]

yes

[lavish portal]

yes

use that

I wanna do this problem too cuz it looks cool

yes it's a very good problem, tell me if you got the answer

[sudden badger]

did u work with the LHS or the RHS or both

im trying to work only with one side of it

[worthy lance]

Lhs

[lavish portal]

lhs yes

[lavish portal]

this rule is all you need

[sudden badger]

i dont know if im doign this right

ill send

[worthy lance]

You are wrong in the last two lines

[sudden badger]

oh yeah

[worthy lance]

What is that

[worthy lance]

Nop

[sudden badger]

oh

2cos

right?

[worthy lance]

Aha and take common factor to simplify

[sudden badger]

yea

do i use double angle formula for cos

[worthy lance]

Do it again for cos^2(2t)

[sudden badger]

huh

do what again

[worthy lance]

.

[sudden badger]

ive simplified the 2 no?

[worthy lance]

Right sorry

[sudden badger]

(cos^2(2t) + 1)/2

i use it there

[worthy lance]

Yes

[lavish portal]

good work, watch out for this tho, you can't set the LHS = to the RHS UNTIL you prove that they are equal

[sudden badger]

i got to

(cos^4(2t))/2

[worthy lance]

?

[lavish portal]

wait what

what rule did you just use

[sudden badger]

after common factor

second line

oh

[lavish portal]

you're going the opposite direction

use this

[worthy lance]

$\cos^2(2t) = \frac{1 + \cos(4t)}{2}$

[jolly parrot]

Samuel

[sudden badger]

(3 + cos(4t)/4

oh its done

thats 3/4 + cos(4t)/4

[lavish portal]

yuayayayayay

vonbrats

congrats

[sudden badger]

thank you

.solved

[pearl pond]

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[pearl pond]

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[brisk coral]

Is there a way to simplify

[brisk coral]

$\sum_{b=1}^a{c\choose b}$

[jolly parrot]

Qwertious

[pearl pond]

@brisk coral Has your question been resolved?

[sharp vigil]

,w sum from b = 1 to a of (c choose b)

[jolly parrot]

Results provided by WolframAlpha

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Requested by cloudedhorizons

[sharp vigil]

not a particularly nice one

[brisk coral]

What does the 2F1 there mean?

[sharp vigil]

hypergeometric function

[brisk coral]

Got it thx

.close

[pearl pond]

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[sudden badger]

I’m trying to simplify this as much as I can

How should I continue

idk if i should half-angle formula for tan because i'd have square roots

[pearl pond]

@sudden badger Has your question been resolved?

[rugged niche]

First step: only simplify tan(a)+tan(b), leave the denominator. Second step: expand the fraction with (1 + cos(a+b)) and use pythagorean identity in the denominator. Third step: tan(x/2) = sin(x) /(1+cos(x))

[sudden badger]

.close

[pearl pond]

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[sharp smelt]

$dim(V_1+(V_2+V_3))= dim(V_1)+ dim(V_2+V_3) - dim(V_1 \cap(V_2+V_3))$
\
\$dim(V_1+V_2+V_3)=dim(V_1)+dim(V_2)+ dim(V_3) - dim(V_2 \cap V_3)-dim(V_1 \cap ( V_2 + V_3))$
\
We now consider $dim((V_1+V_2)+V_3)$ and $dim((V_1 +V_3)+V_2)$ as well ,which are obtained similarly, add them, and divide by 3, to obtain
\
$dim(V_1+V_2+V_3)= dim(V_1) + dim(V_2)+ dim(V_3) - \frac{\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim ( V_2 \cap V_3)}{3} - \frac{dim((V_1+V_2) \cap V_3) + dim((V_1+V_3) \cap V_2)+ dim ((V_2+V_3) \cap V_1)}{3}$

[jolly parrot]

A dense set

[sharp smelt]

Is this too concise?

[pearl pond]

@sharp smelt Has your question been resolved?

[sharp smelt]

<@&286206848099549185>

[pearl pond]

@sharp smelt Has your question been resolved?

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[sharp smelt]

45

[sharp smelt]

To do this, I'd have to find the number of solutions of $5e^{-0.1|x|} sin(x)=1$

[jolly parrot]

A dense set

[sharp smelt]

How would I do that without a calculator

What I can say is that there are a countably finite number of critical numbers, as $e^{-0.1|x|}= \frac{1}{5}$ has exactly two solutions, one less than than zero, and the othr more than $0$

[jolly parrot]

A dense set

[hallow raven]

i’m not seeing how that follows

[sharp smelt]

Basically after $e^{-0.1|x|} attains 1/5$, the amplitude to the sin function beomces less than 1

[jolly parrot]

A dense set

[sharp smelt]

so $5e^{-0.1|x|} sin(x)=1$ isn't possible after a point

[jolly parrot]

A dense set

[grim fractal]

yes so you just need to estimate those values

I think it'd only need to be relative to multiples of 2pi (roughly)

[sharp smelt]

desmos suggests 10 solutions

[hallow raven]

wait sorry, are you saying that there are at most two solutions?

[grim fractal]

yes I also graphed it

[sharp smelt]

I'm saying that there are two points at which $e^{-0.1x}= 1/5$

[jolly parrot]

A dense set

[midnight haven]

change the eq to 5sinx = e^(0.1|x|)

[grim fractal]

But basically, as you were saying, you have two x values for this, so all the waves between these two will have two solutions each

[hallow raven]

ic

[midnight haven]

n check if the maximum of lhs is bigger than rhs

[sharp smelt]

hmm?

Again, I would like to estimate this without a calculator

I guess that's really ahrd

*hard

[midnight haven]

you need to approximate 10ln5

[sharp smelt]

hmm, that will require a calcuator

that;s 17

sorry, 16

[midnight haven]

now compare that with (4n-3)/2 pi

find the smallest n that satisfies (4n-3)/2pi > 10ln5

n is a natural number

[sharp smelt]

why 4n-3/2π?

[midnight haven]

thats when sin(x) reaches maximum value

[sharp smelt]

Ah

so doing this manually is a game of approximations

[midnight haven]

sometimes

[sharp smelt]

In this case

[midnight haven]

im doing this only because i cant find the roots of f''(x)

[sharp smelt]

Yeah, got it

Thanks!

Can I close this now

[midnight haven]

no

keep it open forever

[sharp smelt]

Differentiating f(x) , we find that $f'(x)=ax^{a-1}(1-x)^b+ bx^a(1-x)^{b-1}$.
\
Using fermat's theorm , we know that if a local extremum occurs at $c \in [a,b]$ then $f'(c)=0$.
\
It therefore follows that if $f'(x)=0 \implies ax^{a-1}(1-x)^b+ bx^a(1-x)^{b-1}=0$ . Solving it we find that $ax^{a-1}(1-x)^b = -bx^{a}(1-x)^{b-1}$. From which it follows that $ax^{-1}=-b(1-x)^{-1}$
\
We thus have $a-ax=-bx$
\
Which tells us that $x = \frac{a}{a-b}$ is a solution

[jolly parrot]

A dense set

[spiral pivot]

You have proven a/(a-b) is an extremum, but have you proven it a maximum?

[sharp smelt]

I can't use the double derivative test yet

[spiral pivot]

Don't need it

[sharp smelt]

hmm,at x=0 and x=1, f(x)=0

[spiral pivot]

Yup

What about f(1/2)?

[sharp smelt]

(1/2)^{a+b}

[spiral pivot]

Which is a positive number

[sharp smelt]

yes, now let's check $f(\frac{a}{a-b}$

[jolly parrot]

A dense set

[spiral pivot]

Oh

[sharp smelt]

yes?

[spiral pivot]

Oh nevermind

I'm blind

Please disregard

[sharp smelt]

We then have $(\frac{a}{a-b})^{a}(\frac{-b}{a-b})^b$

[jolly parrot]

A dense set

[sharp smelt]

From this we can conclude the answer depends on the parity of a and b

[spiral pivot]

Chain rule

[sharp smelt]

oops

$ax^{a-1}(1-x)^b=bx^a(1-x)^{b-1}$

[jolly parrot]

A dense set

[sharp smelt]

so $\frac{a}{x}= \frac{b}{1-x}$

[jolly parrot]

A dense set

[sharp smelt]

From this it follows that $x= \frac{a}{a+b}$

[jolly parrot]

A dense set

[spiral pivot]

That looks more reasonable

And now you should always get positive numbers

[sharp smelt]

From which $(\frac{a}{a+b})^{a}(\frac{b}{a+b})^b$

is obtained

Now to prove it's the maxima

[spiral pivot]

You mean a+b

Not a-b

[jolly parrot]

A dense set

[sharp smelt]

Pretty sure to verify this is indeed the maxima , I need the double derivative test

[spiral pivot]

You don't

[sharp smelt]

The converse of fermat's theorm isn't true

[spiral pivot]

You just need Rolle's theorem to show existence, the fact that there are no other roots than 0 and 1 to show that the sign never changes, a single point between the two roots to show that it is positive between the two roots, and finally only the single solution for the extremum

Therefore, the extremum must be a maximum

[sharp smelt]

wait, what

I understand that part about rolle's theorm

but I'm not sure I get how it's a maximum

[spiral pivot]

So this is a continuous function

And it has two zeroes and between them a positive section

And there is only one extremum

[sharp smelt]

Yes, but I'll then have to prove a positve section, with an extremum implies a maxima

On second thought, that's probably what I'd have to do in a RA course

not an introductory calc course

ok yeah

that works

thanks

.close

[pearl pond]

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[agile ridge]

Is there a series Σan where an = o(1/n) but Σ|an| diverges ?

[cinder flower]

what do you think?

[agile ridge]

I think no, afaik all 1/n^p are convergent

but I'm not sure

[west sapphire]

try an alternating series

what's the simplest alternating series you can think of?

[agile ridge]

(-1)^n/n^2

[cinder flower]

hello bungo analysis sir

[west sapphire]

hello generating functionologist ma'am

[west sapphire]

try without the square in the denom

still alternating, does it converge?

[agile ridge]

no

[west sapphire]

close

[agile ridge]

but it's not o(1/n)

[west sapphire]

no?

define o(1/n)

[west sapphire]

i always forget which of these silly o's and O's does what

ah i see

[agile ridge]

yea so it's not

[west sapphire]

ok so if you have n in the denominator then you're not o(1/n)

and if you have n^p for p> 1 then you're gonna converge

so usual idea here would be to try n log(n)

[agile ridge]

why?？ where does n log(n) come from?

[cinder flower]

bungo’s hat

[west sapphire]

1/(n log(n)) is a standard example of something that's o(1/n) but still larger than 1/n^p for any p>1

[agile ridge]

It's diverging?

[west sapphire]

yea

but the alternating version converges

[agile ridge]

so it's conditional converging?

[west sapphire]

yep

[agile ridge]

tysm

.close

[pearl pond]

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[nocturne night]

Ok sorry it's 2am and I have no braincells left

[nocturne night]

84

I legit do not comprehend the question

[pseudo oxide]

what subject is this?

[nocturne night]

Wait

Calculus, parametric/polar

[pseudo oxide]

ahk

[nocturne night]

Wait nvm I'm just being an idiot lmao

.close

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[buoyant garnet]

One message removed from a suspended account.

[buoyant garnet]

One message removed from a suspended account.

One message removed from a suspended account.

[nocturne night]

Lol

Sinx*cscx is just 1 no

[buoyant garnet]

One message removed from a suspended account.

[nocturne night]

Yes(?)

[buoyant garnet]

One message removed from a suspended account.

[nocturne night]

Yeah ...

Sinx*cscx = 1

D/dx of 1 is just x

Nope

Is just 0

Sorry thinking of integrals 💀

[buoyant garnet]

One message removed from a suspended account.

One message removed from a suspended account.

[compact ridge]

no

[nocturne night]

You can simplify the inside first....

[buoyant garnet]

One message removed from a suspended account.

One message removed from a suspended account.

[pearl pond]

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[north talon]

is what i did correct?

[north talon]

the question is in the first two lines btw

[pearl pond]

@north talon Has your question been resolved?

[north talon]

<@&286206848099549185>

[fresh flame]

yes

[snow stratus]

Believe so.

[north talon]

thanks

.close

[pearl pond]

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[crimson bobcat]

hi

[crimson bobcat]

hi everyone

can anyone help me find x

i do know the answer the tho

the answer is 1

but i cant find the solution

i hav tried a million solution

alrdy but nothing seems right

??

can anyone

help me here

[icy narwhal]

damn

ok

lemme think

[crimson bobcat]

like

the solution is not simple

[icy narwhal]

nah it aint

[crimson bobcat]

like adding

[icy narwhal]

you got this in a logic class ?

[crimson bobcat]

multiplying

doesnt

work at all

[crimson bobcat]

olympiad

class

[icy narwhal]

ahah

not ez

[crimson bobcat]

ik

[icy narwhal]

OK I GOT IT

IM A GENIUS

@crimson bobcat

[crimson bobcat]

yes

how

[icy narwhal]

the sum

of one side

ok wait

[crimson bobcat]

it didnt work

i hav tried it

since the start

not a single side

[icy narwhal]

you

[crimson bobcat]

work

[icy narwhal]

ok

look

for example

[crimson bobcat]

yuh?

[icy narwhal]

yes

wait

idk how to explain typing

go dm

[pearl pond]

@crimson bobcat Has your question been resolved?

[crimson bobcat]

welp welp

welp

<@&286206848099549185>

can anyone help

hihi

can anyone hlep me with

thethe circle thing

can anyone hel pme find x

hi

can anyone hlep me

find htis one

[pearl pond]

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[woven bluff]

what was the solution? 🤔

[pearl pond]

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[latent quail]

Is it always required to indicates | x - c | is greater than zero in the last line?

[pearl pond]

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[latent quail]

.close

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[midnight haven]

😭

[wanton cairn]

ye bro you alone in this one 🙏🏻

No jk, I didn't study this type of integrals for the moment

like wtf in + inf

[sinful nebula]

sub

[midnight haven]

sub what

[sinful nebula]

$\int_{0}^{\infty }\frac{ln(x)}{x^{2}+k^{2}}$

[jolly parrot]

Alaska

[midnight haven]

i was thinking ibp lol

[sinful nebula]

theta = 1/k arctan(x/k)

[midnight haven]

ans; x = 2 tan(theta)

[sinful nebula]

$\Theta = \frac{1}{k}arctan(\frac{x}{k})$

[jolly parrot]

Alaska

[sinful nebula]

$x = k*tan(k\Theta)$

[jolly parrot]

Alaska

[sinful nebula]

now its not that hard

[quick star]

i think you can substitute x = e^y and exploit some symmetry conditions

[pearl pond]

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[dim nest]

if $Q$ is a rectangular matrix such that $Q^TQ = I$, then I managed to show that
$|QA| = |A|$ i.e. it preserves the norm [spectral norm] \

is it also true that $|Q^TA| = |A|$ ? i've been trying to prove it unsuccessfully, so now i'm thinking it might not be true

[jolly parrot]

ego

[pearl pond]

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@dim nest Has your question been resolved?

[dim nest]

it is not true, managed to find a counter example 🙂

.close

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[tall fog]

How do you get this???

[tall fog]

Apparently it's correct but where the hell did that come from

The second part, the multiplication is what I dont understand. I get it came from the lower part of the first part somehow, but how??

[jolly parrot]

faiyrose

[tall fog]

Ahh but I dont understand how the formula is applied. I cant grasp it for some reason

What do you mean?

Sorry if it's obvious I just dont see it

I guess a=3√2 and b=1

[jolly parrot]

faiyrose

[tall fog]

So why is only the middle part put in?

What about the (3√2+1) left of it? Just ignored or?

[jolly parrot]

faiyrose

[tall fog]

(a+b)(a²-ab+b²)=a³+b³ if that's what youre asking

Oh wait

Is it because we already have the left side ( a and b ) so we just now multiply it with the right side that we don't have?

Oh alright, I get it now! Thanks

.close

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[pearl pond]

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[untold dust]

In a culture containing $510^8$ bacterial cells and $310^8$ bacteriophage, what proportion of the bacterial cells would have three or more phage particles absorbed to their cell walls?\
Can comeone help me with figuring out which distribution I need to use?

[uneven kindle]

I think i miss some context here

[untold dust]

This is the entire prompt
I need to figure out which statistical test/distribution I need to use and explain why

[midnight haven]

Use poisson distribution

[jolly parrot]

KySquared

[untold dust]

Are you sure?
Like I kinda understand the logic since bacteria growth is a continuous variable but I don't understand why poisson specifically

@midnight haven

[midnight haven]

Okay well like

I feel that this is applicable because you are looking at the number of phage particles absorbed per bacterial cell, which can be treated as a rare event in a large population in a way

[untold dust]

But then I would need the sample mean, no?

[midnight haven]

Why sample mean?

[untold dust]

Nvm,
I realized 5*10^8 is the population

[midnight haven]

Yeah

[untold dust]

so mu = 5*10^8 and x=3?

[untold dust]

I'm not sue this is right

I threw the poisson distribution at it with those parameters and I got 0

(makes sense since its e^(-10^8)

[pearl pond]

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[fresh silo]

hello

[fresh silo]

need help for this

i need to identify the conic

and find the focal point and the summit

[midnight haven]

Re arrange the equation

[fresh silo]

wdym by re arranging

[midnight haven]

Are u identifying the conic

Re arrange equation to y²= x²-2x+2

And complete the square for the x terms

[fresh silo]

y²=(x-1)²+1 that's what you mean ?

[midnight haven]

Yes

[fresh silo]

then what i do

<@&286206848099549185>

@midnight haven hey

[midnight haven]

Hi

[fresh silo]

im lost

[midnight haven]

Okay so

With this equation

U identify the vertez

Vertez

Oml

VERTEX

[fresh silo]

wtf is vertex

ah the summit

i need to find the vertex of (x-1)²+1 ? @midnight haven

idk how to find it

[midnight haven]

Oh wait

I think

U can

Hold on

Rewrite the equation

[fresh silo]

y²=(x-1)²+1 ?

[midnight haven]

U write that into standard form

So it becomes

[fresh silo]

y²=x²-2x+2

[fresh silo]

what i can do with this

[midnight haven]

U re arrange it

Into

Uh

[fresh silo]

it cant be this

a and b need to be different

[midnight haven]

Ack

[fresh silo]

false

[midnight haven]

My brain hurts 😭

[fresh silo]

wait

this are the answers

vertex are (1,1) and (1,-1)

[midnight haven]

OH

OH

OH

AAAAA

[fresh silo]

focal point are (1, sqrt(2) etc

maybe it something with y²-2 ?

idk

[midnight haven]

Hmm

Try it

[fresh silo]

idk 😂 we maybe need to do something like (y-sqrt(2))(y+sqrt(2))

[midnight haven]

😭😭 TRY IT

[fresh silo]

im bad ...

no wait

hmm

if the focal point is sqrt(2)

that means they did sqrt(1²+1²)

but a and b cant be 1 they need to be different

the only solution possible is that a and b are equal to 1 but its not possible

you understand what i mean

c²=a²+b²

c=sqrt(a²+b²)

[fresh silo]

and here the "c" is sqrt(2)

that means a and b are equal to 1

is this possible @midnight haven ?

oh

a can be equal to sqrt(3) and b to 1