help-13

[rapid mauve]

logic?

[mellow thistle]

same but with days

60 x 60 x 24 x 365

[rapid mauve]

there we go.

[mellow thistle]

ahhh

okay thank you

[rapid mauve]

no trolling in the future, i hope.

not when someone is trying their best to help you

please respect the effort of your helpers

not this problem, but i think you remember what you did with me ytd

aight, well done

[low edge]

Good job to everyone who contributed

[low edge]

you can type .close if you're done

[cedar kiln]

@mellow thistle Has your question been resolved?

[mellow thistle]

How did this happened?

[low edge]

Hey! Is this an unrelated question to your original one that you posted in this thread?

[cedar kiln]

@mellow thistle Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[barren hemlock]

can someone check my proof since it's really wonky and i've been solving for 7 straight hours now

[barren hemlock]

and if it's not clear i used am-gm on the first inequality at 5^z+3^t

btw no need to check after x=2 or y=2 since i kinda gave up on the write up ill fix that mess later i just need to know if it works works

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[abstract breach]

Is there any way to prove this without complex analysis?

[nocturne ravine]

Mais non Thomas

Dingue

[abstract breach]

mdr

we only need to look at polynomials of the form $X-\alpha$ with $\alpha \in C$

[wraith dagger]

bloubbloub

[tropic merlin]

I FINNALY FOUND THE PLACE WHERE I BELONG

[abstract breach]

when $|\alpha| \neq 1$ this can be done with riemann sums (?), but how to do it when $|\alpha|=1$?

[wraith dagger]

bloubbloub

[tropic merlin]

oh sorry this is help

[abstract breach]

$\exp(\int_0^1 \ln{|e^{2i\pi t} - \alpha|}dt), |\alpha| = 1$

[wraith dagger]

bloubbloub

[dire geode]

No

[abstract breach]

crazy

.close

[cedar kiln]

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[stiff coral]

Hello, i dont have an specfic exercise to ask. I have a concept problem. Im studying for my complex analysis exam, and i dont understand what is the difference between a pole and a singularity, can someone enlight me? thanks

[flat mica]

a pole is a type of singularity; all poles are singularities

[stiff coral]

Ok but why not all singularities are not poles?

[flat mica]

three types of singularities

removable singularities, such as, sin(x)/x

poles, which you hopefully know

and essential singularities, where lim z->0 f(1/z) and lim z->0 1/f(1/z) both fail to exist

[stiff coral]

so a pole is a removable singularity? idk how is called in english

[flat mica]

no

removable singularities are not poles

[stiff coral]

i still dont get it sorry

[dreamy void]

When you have 1/z you cannot remove z=0

Removable would be z/z for example

Or sin(z)/z because of the power series expansion of sine

[stiff coral]

yes 1/z has a escential singularity at 0 no?

and a pole at 0 too

[dreamy void]

No

An essential singularity is given if the Laurent series has infinite many terms with negative powers

The laurent series of 1/z is just 1/z so it's only a pole of order 1

What would be an essential singularity would be sin(1/z)

As a rule of thumb. when it comes to rational functions without common roots of numerator and denomintaor, then only poles come into question as singularities

[stiff coral]

Okay that helps

i still dont fully grasp it bc we didnt see series yet

but the rule of thumb helps a lot

i thing the confusion is more on escential singularities more than poles ig

think*

[dreamy void]

The difference between a pole and essential singularity is, when it comes to a pole of f then the limit of |f| -> oo. That is not the case with essential singularities, because they are unpredictable in a sense where the limit can take any value from different routes

https://en.wikipedia.org/wiki/Essential_singularity#:~:text=For essential singularities of real,do not have a residue.

You might wanna check out the alternative descriptions of singularities and how to spot them

[stiff coral]

alright will do, thanks

complex infinity is wierd man

.close

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[zealous palm]

stuck here

[surreal cave]

(x\ln6-2x\ln6=-x\ln6)

[slender ginkgo]

any common factor

that u notice?

[zealous palm]

factored x

[buoyant latch]

You don’t have to ln both sides

aˣ is injective for positive a

[zealous palm]

omg

i’m so dumb

[slender ginkgo]

can u simplify the rhs

[zealous palm]

i could have just only done the exponents

yea i see im so slow for not noticing

[buoyant latch]

Yes

But this is only true because of the injectivity thing

[zealous palm]

ignore my hand writing

i wrote it angry

[buoyant latch]

Yeah

Haha

It’s alright to not see it, it comes with practice!

Sometimes it’s obvious but you miss it it happens

[zealous palm]

my exam is in 2 hours

not much practice left lol

[buoyant latch]

Your maths education does not stop after your exam

[zealous palm]

oh yes it does😭😭😭

this is my last math class for the rest of my life

besides physics but ts easy

and i won’t be using imaginary garbage

[buoyant latch]

Yeah well that’s not the right outlook to maths, or any other subject

[zealous palm]

this is not part of my degree i don’t give a single fuck about this class

or math

[buoyant latch]

(Unless you study anything that has to do with physics or engineering)

[zealous palm]

i study kinesiology

so like human physics

but it’s rarely like that i feel like

should i use log or ln for these

[buoyant latch]

Learning maths is more about the methodology and mindset than the actual mathematics for ordinary people

[slender ginkgo]

Oh

[zealous palm]

i find it easy to learn most of the time but the work load of this class with my schedule is creating crazy burn out

i wish i could take 2 days off

[buoyant latch]

If you didn’t realise that then your institution has failed ya :/

[buoyant latch]

Yeah tbh it’s rough like that sometimes

[zealous palm]

it’s american education they just want factory working drones

it’s never been about educating the public

[buoyant latch]

;c

[zealous palm]

that’s why it will never be redesigned

it was corrupt from the beginning

they want you dumb fat and poor

that’s how they make money

mb ill go back to math😭

[zealous palm]

so ln or log

[buoyant latch]

These are cringe

Because log 5 or 8 both work only on 1 side

[zealous palm]

can i just double it bc it’s 5 and then use log to cancel it

log10

x2

and then square root the end

[buoyant latch]

I would say use log base 5 then change of base on the 8 or vice versa

5² is not 10 it’s 25

[zealous palm]

no im saying

i would be left with x2

[buoyant latch]

No you can’t do that

[zealous palm]

o

[buoyant latch]

$\log_5(a) \neq \sqrt{\log_{5*2}(a)}$

[wraith dagger]

frosst

[zealous palm]

huh

[buoyant latch]

What

Where did 10^2x come from

[zealous palm]

double 5^x😭

[buoyant latch]

But 5^x isn’t the same as 10^2x

I mean plug in x = 1 and you get 5 = 100

But I think I know what you’re thinking

Consider the problem $8^{3x + 1} = 2^{x-1}$

[wraith dagger]

frosst

[buoyant latch]

You can here turn $8^{3x + 1}$ into $(2^3)^{3x+1} = 2^{3(3x+1)} = 2^{9x+3}$

[wraith dagger]

frosst

[zealous palm]

wtf

i moved on my brain hurts and i feel like i can kinda do them

do i negate the log3 and then distribute x+2 into d+8

[buoyant latch]

Don’t “negate”

[zealous palm]

that’s wat the last one did

[buoyant latch]

Take 3 to the power of both sides

Then you have 3^(a + b) = 3^a * 3^b where a and b are the log stuff

[buoyant latch]

The word “negate” is ambiguous

[buoyant latch]

This is clear

[zealous palm]

this is the last one

log7 got deleted

[buoyant latch]

Why is it equal to = 0

x² - x = 6 = 0?

[zealous palm]

the next step was to do x^2 - x - 6 = 0

so i just skipped moving it

ik what i mean

[buoyant latch]

This only happened because what you really did was take 7^both sides

$\log_a(b) = \log_a(c) \implies a^{\log_a(b)} = a^{\log_a(c)}$

[wraith dagger]

frosst

[zealous palm]

yea i’m bad at applying these

that’s why i write out the steps and then replicate them

[buoyant latch]

Yes but there’s rhyme and reason to why these things work

[zealous palm]

i understand that but i don’t have time to learn them

or how they apply

[buoyant latch]

So for example, $\log_2(x) = \log_3(2x)$

[wraith dagger]

frosst

[buoyant latch]

This you can’t just “negate” the logs

[zealous palm]

how do i rewrite the original equation

[buoyant latch]

Which one

[zealous palm]

log3

[buoyant latch]

This one?

[zealous palm]

this

[buoyant latch]

Or this one

[buoyant latch]

You can first multiply both of the logs together

log3(x² + 10x + 16) = 3

Then take 3^ both sides

That gets you x² + 10x + 16 = 3³

The base 3 and the log on the left cancel out

[zealous palm]

why 3^3

[buoyant latch]

Because of this

$\log_a(b) = c \implies a^{\log_a(b)} = a^c$

[wraith dagger]

frosst

[zealous palm]

so once i have the two x numbers

do i plug them into the original equation

and if it’s negative i can ignore it

[buoyant latch]

If the x would make either x+2 or x+8 non-positive then ignore it yeah

You can’t put negative or 0 inside ln so any x that makes that happen breaks the whole thing

[buoyant latch]

Notice that these implications are 1 way, just because your quadratic says x = bla bla bla doesn’t mean it’s a solution to the original equation

[zealous palm]

so i redo the entire equation with those as x

[buoyant latch]

But if it is a solution then we know it must be a solution to the quadratic

[buoyant latch]

No you just have to look at x + 2 and x + 8

See if they are non positive and throw out the solutions that make it non positive

[zealous palm]

ok

[buoyant latch]

Whatever is inside ln can’t be non positive

So the x can’t make the inside non positive or it’s invalid

[zealous palm]

this seems so easy to fuck up

if this was my degree i’d kill myself omg😭

i’ll never understand how people do ts

[steady ermine]

"Math is easy, arithmetic is hard."

[zealous palm]

you got some crazy spread sheet takes

why did it turn into 6x

[steady ermine]

x*(x+6)=x^2+6x, agreed?

[buoyant latch]

Multiply

It’s just distribution

[zealous palm]

oh i added the x’s😾

this is what i mean i miss one step and my answer way off

i feel like a 5th grader😭

[steady ermine]

That's why it's always important to show your work.

(even at the professional level, within reason)

[zealous palm]

do i just solve

will my calc read -e^6

[steady ermine]

If x=2-exp(6) is the solution, you leave it at that (as you cannot simplify).

Not sure how you got there, though.

[zealous palm]

it’s all written??😭

[steady ermine]

You started at ln(2-x)=6?

[zealous palm]

yes

[steady ermine]

If so, then that's fine.

[zealous palm]

no

ln(-x+2)=6

[steady ermine]

-x + 2 == 2 - x

but yeah, you're good as exp(6) isn't rational.

[zealous palm]

so what is x

[steady ermine]

You wrote that it's equal to 2-exp(6).

It is what it is.

[zealous palm]

i can only answer in alternate form

how do i get from solution to alternate

[steady ermine]

Calculator

[zealous palm]

it didn’t work

[steady ermine]

What exactly didn't work?

[zealous palm]

nvm my phone calc is just downy i guess

it worked on my test calc

[steady ermine]

And that's what's important, lol.

but might want to make sure you know how to do it on your phone, as that's what you'll realistically have on you irl.

[zealous palm]

i’m never ever doing this ever irl

i would bet every dollar i make my whole career

[steady ermine]

Eh, don't knock it. Everything's hard at first, maybe it gets better?

Unless you already know what you want to do, then prosper.

[zealous palm]

i do

i’m halfway through my degree

[steady ermine]

Ah, so Junior?

[zealous palm]

yea

[steady ermine]

Sounds good, I've known those who'd graduate and go back as an undergrad instead of continuing as a grad.

Anyways, gl and gw

[zealous palm]

can i js ignore the ln or do they effect the problem

[steady ermine]

In which step?

[zealous palm]

all of them

[steady ermine]

Are you asking if you can drop them, or if you can replace them with something else like log or lb?

[zealous palm]

drop them entirely

they annoying and seem pointless besides multiplying

unless that isnt valid with more complex

[steady ermine]

Well then you'd have ln(4) + ln(4+1) = ln(20) implying that 4 + 4 + 1 = 20, which is clearly a contradiction.

[zealous palm]

very slightly more complex mind oyu

nno but its in parentheses so i would multiply

(4)(5)

=20

[steady ermine]

Oh you mean when you have ln(4*5)=ln(20)

[zealous palm]

this exam is going to be a fucking trainwreck

[steady ermine]

The reason it works to drop it off then is that you can raise both sides by the power of exp

so exp(ln(4*5)) = exp(ln(20)) ==> 4*5 = 20

as exp and ln are inverses.

[cedar kiln]

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[sturdy escarp]

can someone please real quick explain where these are derived from

the reduction formulas

I needa memorize sin cos tan and sec

in like a couple hours

or I just derive it during exam

[tranquil oracle]

Main idea is realizing sin^2 + cos^2 identity to remove 2 powers of 2. Also, can integrate by parts (cos x)(sin^(n-1) x cos x)

Similar for cos

For tan, use sec^2-tan^2 identity and you are pretty much done

You can practice deriving it in the few hours

[gentle echo]

the

integratin by parts?

[tranquil oracle]

For sec, i think besides using the identity, you need to integration by parts

[sturdy escarp]

can I just use these identities and then just u-sub?

do these formulas have any use other than being more efficient?

might just be quicker to fully do it by hand than try to recall these formulas or fully derive on an exam

[tranquil oracle]

Seems rather unlikely

[gentle echo]

well you have a series of descending powers, that's useful for

say like sin^7(x)

[tranquil oracle]

Yeah just practice a few, you need to get used to integration by parts

I recommend at least doing sin, tan and sec, since cos would be similar to sin

if you want, skip tan and do sec because tan is easier and doesnt need integration by parts

[sturdy escarp]

hmmm

yeah okay ig I'll just learn them

tysm both of you

.close

[cedar kiln]

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[tranquil oracle]

Oh yeah you definitely need a u-sub during your integration by parts

I just implicitly thought it

[cedar kiln]

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[runic dock]

birthday paradox

[runic dock]

i want to know how to calculate how many people are needed to reach a certain probability that two people share a birthday

the expression simplifies to

(365-1) * (365-2) * ... * (365-n) < p * 365^n

where n is the number of people and p is the probability that two share a birthday

so lets say i want to know how many people needed to reach 50%, which gives (365-1) * (365-2) * ... * (365-n) < 0.5 * 365^n, solve for n

thanks in advance

[tropic oxide]

theres no clean way to do it without like trial and error

[runic dock]

ic ty

.close

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[crimson sedge]

wait can anyone make sense of this powerpoint?

[crimson sedge]

just don't understand what it's trying to teach... or the usefullness of learning it

[opal hinge]

!ss

[cedar kiln]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[crimson sedge]

is it keynote 😭

this one might be better

no???

bruh

[rapid mauve]

take screenshots

[teal wedge]

no one wants to download

[tropic oxide]

my brother in christ what the hell is a .key?

[earnest socket]

Apple Keynote file?

essentially the Apple version of Powerpoint

[cedar kiln]

@crimson sedge Has your question been resolved?

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[crimson sedge]

im not exactly sure where i made a mistake my final expression miss the y i canceled the y from the second term is it wrong to do that

[haughty osprey]

Don't cancel (x-8) in the second term , instead multiply and divide the first term with xy

[crimson sedge]

wait so instead of canceling (x-8) we just multiply xy to first term to match it?

[haughty osprey]

Yes

[crimson sedge]

i got the right answer

thanks :)

.close

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[keen geyser]

I have a question regarding complex analysis, given that f(z) is holomorphic, under what condition is |f(z)| also analytic?

[tropic oxide]

if it's constant lmao

a purely real function cannot be holomorphic without being constant

[keen geyser]

hm yes

I am thinking of whether or not I can apply lagrange mean value theorem to this type of modulus function

For example if I have a nonconstant holomorphic function f(z) and |f(z)| is constant about some contour say its constant on the unit circle. I want to reach the conclusion where there must be a stationary point within the unit circle for the function f(z)

[tropic oxide]

lagrange mean value theorem?

isn't that for R -> R?

what are you really trying to do here?

!xy

[cedar kiln]

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.

[keen geyser]

There is no original question

Its just something i've been thinking of

[keen geyser]

well not exactly its just something that popped into my head when I think of it

Lemme actually pull a graph out

Ok you see this domain coloring graph of z^4+2

There is a level curve for modulus around the 4 roots indicating that all the points on the curve has the same modulus

So I was thinking since all points on this closed contour has the same modulus does it mean there must be a stationary point within this closed contour

This is a bit similar to the lagrange mean value theorem, maybe theres another theorem that can prove this but idk

[tropic oxide]

so you want to show that if f is analytic and |f| is constant on some contour \gamma then there exists a point inside \gamma where f' = 0

which im not entirely convinced is true but i also cannot produce a counterexample off the cuff.

[keen geyser]

yes

It does really make sense if you consider the 3d domain coloring graph

If we take the height as the modulus

since the analytic function must be continueous its pretty obvious that theres gonna be a peak/trough within right

[tropic oxide]

f'(z) = 0 is not really visible on a modulus graph though is it

[keen geyser]

Its not

yes

Its gonna be visible on 3d graphs

Its just some flat place with constant color

[cedar kiln]

@keen geyser Has your question been resolved?

[keen geyser]

<@&286206848099549185> plz save me

[cedar kiln]

@keen geyser Has your question been resolved?

[earnest girder]

Yo

[cedar kiln]

@keen geyser Has your question been resolved?

[keen geyser]

Plz someone save me

[hallow pelican]

you will be saved if ur called crowd fighter

lone fighter will fight alone.

[spring forum]

@keen geyser you can continue to wait for people who know complex analysis here, or you can go to #real-complex-analysis . you might find more luck there, because people who're proficient in that area tend to be more active there.

[limber monolith]

Hello

[spring forum]

hi, check #discussion for general chatting

[cedar kiln]

@keen geyser Has your question been resolved?

[keen geyser]

not like this

[viscid thunder]

i really need help guys, im having a test tmr
supposedly speaking if i had two planes that were perpendiculared on the same line, would the distance between the two opposite apexes to the two opposite planes equal each other
english is not my first language so this could be a little hard to understand
hopefully you guys can help me

[dire geode]

#study-discussion or open an available help channel with a math problem

[cedar kiln]

@keen geyser Has your question been resolved?

[dapper laurel]

fuh no

[loud oasis]

wtf is this long ah chat

you guys solving the reiman hypothesis in here

[fierce minnow]

Isnt that hard

like really hard

[meager raft]

Nah we solving the Navier Stokes Problem

[cedar kiln]

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[lament hemlock]

how to get help

[opal hinge]

post your problem here and others will try to help

[lament hemlock]

in in hs doing the highest level of math so im not rlly undergrad but the topics in highschool role on this dc server are too basic

[lament hemlock]

i need help with part ii

what partial fraction decomposition should i use?

i tried this

I used this partial fraction to expand the series but i dont notice anything significant after than

[opal hinge]

your role doesn't matter

[lament hemlock]

oh ok ill switch it then

[tranquil oracle]

Okay there's a trick but partial fractioning into what you did also works

Think of it as
1/2 ... -1 ... + 1/2...

[lament hemlock]

reaaly where do i go after using my partial fractioning?

[lament hemlock]

yeah ok

[tranquil oracle]

So you need to align all the terms properly to force cancellation

[lament hemlock]

yep thats what i treid to do

[lament hemlock]

i got this

when expaning the first few terms

[tranquil oracle]

Dont sum yet, just write all the terms in a grid
Interpret the kth term as
+1/2 of 1/(k-1)
-1 of 1/k
+1/2 of 1/(k+1)

[lament hemlock]

mm ok

[tranquil oracle]

So ask: for each k, how much 1/k are you left with?

[opal hinge]

$A=\frac{1}{2(k-1)}-\frac{1}{k}+\frac{1}{2(k+1)} \Rightarrow 2A=\frac{1}{(k-1)}-\frac{1}{k}+\frac{1}{(k+1)}--\frac{1}{k}$

[wraith dagger]

Alexis_Fx

[lament hemlock]

yeah why double minus?

[tranquil oracle]

(After this is done we can go through the fast approach used by math olympiad students)

[lament hemlock]

sure

[dreamy void]

Looks like a nice way of telescoping

[lament hemlock]

bro i get nothing when i telescope

[tranquil oracle]

@lament hemlock think of this interpretation to help you telescope

[lament hemlock]

oh ok

[tranquil oracle]

You see that 1/2 -1 + 1/2 = 0
That means on average each summand "contributes 0 terms to the sum"

[lament hemlock]

right

[tranquil oracle]

So write out the terms systematically

Only cancel terms of the same 1/k type

[tranquil oracle]

nvm

[tranquil oracle]

You might need a few more to see the pattern

[lament hemlock]

is there not a pattern here like the first term increasing by 2 ub tge denominaotr and then the second by 1 and the third by 2?

[lament hemlock]

ok

i can show a different solution but i have no idea whats going on

its the answers to the paper

[tranquil oracle]

I would recommend treating the 1/2 as a coefficient, splitting it from the 1/k

[lament hemlock]

does this make any sense?

[tranquil oracle]

Yeah this is kinda the fast trick

[lament hemlock]

ok and is there any way to know to use that partial fraction?

cause idk how your meant to know to only partially decompose it

[tranquil oracle]

Theres a general sequence of tricks this trick is part of

If you have a rising product or rising quotient

[lament hemlock]

ok so it just comes with prac

[tranquil oracle]

Here its k-1, k, k+1

But lets go through the standard method so that you learn a standard tool

[tranquil oracle]

Write out the terms like this

[lament hemlock]

yeah ok

[tranquil oracle]

Do you see common terms?

[lament hemlock]

gimme sec

[tranquil oracle]

1/2 1/2  1/2 1/3  1/2 1/4  1/2 1/5
-1 1/3   -1 1/4   -1 1/5   -1 1/6
1/2 1/4  1/2 1/5  1/2 1/6  1/2 1/7

Do you see the common terms that can cancel?

[lament hemlock]

like this?

sorry for bad handwriting

[tranquil oracle]

Focus on the terms which involve 1/5

For instance

[lament hemlock]

yeah its kinda like diagonal cancelation

[tranquil oracle]

Do you see how to finish the method?

[lament hemlock]

so youd try generalise right?

[tranquil oracle]

Yeah you can generalize

[lament hemlock]

oh

is it this

1/4 - 1/2n + 1/2n

[tranquil oracle]

Careful

Check carefully which terms survive the cancellation

[lament hemlock]

wait nv

yeah what am i saying

ohhhh

wauit

its just 1/12 right?

[tranquil oracle]

So you see three of the terms at the top left corner survive, right?

[lament hemlock]

liekl this?

[tranquil oracle]

Pretty much

[lament hemlock]

ohhhh ok tysm

[tranquil oracle]

Now that you get the normal method here comes fast approach

[lament hemlock]

so should i approach telescoping like this?

[tranquil oracle]

Yeah you need to look out for things that can cancel

Sometimes it splits into more terms

So fast approach:

${}$
\begin{align*}
\frac{1}{(k-1)k(k+1)}&=\frac{2}{2(k-1)k(k+1)}\&=\frac{(k+1)-(k-1)}{2(k-1)k(k+1)}\&=\frac{1}{2(k-1)k}-\frac{1}{2k(k+1)}\end{align*}

[lament hemlock]

right and is the the reason why you would try this specific partial fraction?

[wraith dagger]

Element118

[tranquil oracle]

Yeah it is

You can try generalizing this approach

[lament hemlock]

yep

[tranquil oracle]

A modification works on summing up (k-1)k(k+1)

[lament hemlock]

oh wow

this way is much simpler

yeah i see it

only one term survives and everything cancels out

[tranquil oracle]

(Hint: consider (k-1)k(k+1)[(k+2)-(k-2)])

And yeah that's the general trick

[lament hemlock]

yeah tysm you were so helpful

i have a final highschool math exam tomorrow

[tranquil oracle]

In fact with this trick, you can also try your hand at approximating sum of inverse cubes

There's the error term when you did the approximation - what if we used the same trick to sum the error terms?

[lament hemlock]

ok im ngl i dont know what inverse cubes are

[tranquil oracle]

The question asked you to sum 1/3^3 ... 1/n^3

Inverse of cubes

[lament hemlock]

oh ok sorry i didnt know the naming convention

makes sense

[tranquil oracle]

You have the upper bound of 1/12

[lament hemlock]

right

[tranquil oracle]

But you can tighten it if you calculate the errors you introduced and try to sum that

Possible but might be tedious

[lament hemlock]

yeah ok thats interesting

just a quick quetsion outta curiousity could you do it with integrals?

to setup that inequiality that is sn < 1/12

[tranquil oracle]

Approximate with integrals? Perhaps

[lament hemlock]

oh ok

[tranquil oracle]

Maybe you need some effort to set up the right integral

[lament hemlock]

could you not inegrate the inverse cubic fucntion?

and take the upper bound as infinite

and lower at n = 3

[tranquil oracle]

How is that bigger than the sum?

[lament hemlock]

wait yeah nvm that wouldnt help prove the sum

[tranquil oracle]

Yeah you might need to change your bound

[lament hemlock]

ok forgot what i said

[tranquil oracle]

If lower is n=2 you should be able to prove

[lament hemlock]

oh yeah true that makes sense

[tranquil oracle]

But you only get a bound of 1/4 if i integrated correctly

[lament hemlock]

oh what

[tranquil oracle]

So you need to choose a better lower bound or bound the error

[lament hemlock]

yeah ok nvm on the integration part then

also do yk how i should approach this?

is it cases?

[tranquil oracle]

Probably some ordering and bounding

Yeah

That's the hint - suppose a is biggest (or tied for biggest)

If you still need help you can open another channel and mention this hint of ordering

[Then people would know where to find the question]

[lament hemlock]

oh alr anyway thankyou for youre help

[tranquil oracle]

[.close to close]

[lament hemlock]

.close

[cedar kiln]

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[royal flint]

i dont get why its not t=m/38 cuz just becuz he traveled the distance and returned doesnt mean the time and distance is being untraveled? like i dont get why they do *2

[upper laurel]

distance cannot be untraveled, youre thinking of displacement

[royal flint]

yea but then whys there *2

am i tweakin 😭

[tropic oxide]

you take some time to get there and then take the same time to go back, so the time for the return trip is double that of one way

so yes you are tweaking most likely

[royal flint]

but its still m/38 ? like the rate is m/38

[tropic oxide]

m/38 is the one-way time

[upper laurel]

say m = 38

[royal flint]

oh wait

i get it

[tropic oxide]

you want to find the time taken to travel BOTH ways

[royal flint]

dawg

i was tweaking

[upper laurel]

thats where the * 2 is hiding

you were thinking of displacement

[royal flint]

i need to double the time

😭

thanks yall :)

.close

[cedar kiln]

Channel closed

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[upper laurel]

np

[warm crystal]

@royal flint

[cedar kiln]

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[royal flint]

im not sure how 90 degrees to the right will change trajectory i think its like \ from B

[sharp spade]

please help me understand quadratic foormulas

[royal flint]

but im not sure

[upper laurel]

it would be southeast, yes

that youre moving east in the first place should tell you enough to choose an answer

[royal flint]

is it like A / B \ C?

[mighty shuttle]

Have you drawn a sketch?

[royal flint]

yea

[mighty shuttle]

!show

[cedar kiln]

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

[upper laurel]

yes

[royal flint]

i found BC i was tryna use pythagoras to find C's coordinate

[upper laurel]

you do not have enough information to find C

[royal flint]

yea

[upper laurel]

you should instead rule out multiple choice answers

knowing the jogger moved east from B, which answers cannot be true?

[royal flint]

well only C make sense

[upper laurel]

and C is correct

[royal flint]

cuz x>3 and y<9

[upper laurel]

yea

[royal flint]

i get it

thanks :)

.close

[cedar kiln]

Channel closed

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[upper laurel]

np

[cedar kiln]

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Send your question here to claim the channel.

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

Hi! I'm doing some pre-calc trig rn and I am struggling with well everything but I need help with the law of cosines. What I am looking for is a step guide and then I would like to follow the guide with a different problem and someone fact check me. I'll attach a ss and if someone could just walk me through that would be sigma.
If ​​ a=16.2, b=13.2, b=13.2​, and c=17.7. Solve for the angle B.

[chrome quail]

Cosine law

Oh whoops didn’t see you mentioned it

One sec

c^2 = a^2 + b^2 - 2ab cos(C)

It’s pretty much set up, you plug in a,b,c and solve for cos(C), which represents the angle opposite to side c

Get cos^-1 of your answer and that’s all

[neat cave]

something like the formula of resultant of vectors

?

[neat cave]

how did u get that formula

[chrome quail]

It’s cosine law

[thick flint]

Yeah except you should switch some of the variables around cuz they are looking for B

[chrome quail]

Yeah ^

[thick flint]

b^2 = a^2 + c^2 - 2ac cos(B)

[potent hazel]

okay so im filling out the values and this lowkey might be a rlly dumb question but cos(B) is being multiplied right

[lone dune]

yes

no question is a dumb question when you're learning

[thick flint]

Yeah

[potent hazel]

first thank you and second, I put it into desmos and got 113.57

is this correct?

[thick flint]

It might be easier if you isolate cos(B) before plugging things in

[thick flint]

I dont think so

Can you show your work

[potent hazel]

I just plugged it into desmos but I just checked if it was in radians and it was... it's supposed to be in degrees isnt it

Doing that changes it to 17.4

Does that sound better?

[chrome quail]

but you also forgot b^2

[tropic oxide]

heres a tip

[chrome quail]

Woah wait you're way off

[tropic oxide]

you can define variables in desmos

[chrome quail]

You put a distance value (b = 13.2) as an argument of cosine

cos(B) is what you're trying to solve for

[tropic oxide]

that also

[thick flint]

This is what b^2 equals

[tropic oxide]

you can literally put on separate lines:
a = 16.2
b = 13.2
c = 17.7
and then type the formula for what you want on a fourth line

[potent hazel]

I'll do this

[chrome quail]

May as well learn to use desmos properly if you're using it instead of a calculator

[potent hazel]

but also I thought I could just plug in the values?

[thick flint]

Mmm

[potent hazel]

B^2 =a^2+c^2-2ac cos(B)?

[chrome quail]

You could, but declaring values gives modularity, and lets you work with different a,b,c values

[thick flint]

b^2 = a^2 + c^2 - 2ac cos(B)

b and B are not the same

[tropic oxide]

careful not to confuse uppercase with lowercase

[chrome quail]

b^2 =a^2+c^2-2ac cos(B), where lowercase is distance, and uppercase is angle

[tropic oxide]

also yeah as people have said you're solving for an angle not for a side, so

[chrome quail]

Honestly if this is the first question you're doing, i would solve it with pen, paper and a regular calculator

Instead of playing with desmos

[thick flint]

What you need to solve for is cos(B) = (b^2 - a^2 - c^2)/(-2ac)

[potent hazel]

Yeah i have paper lol, i'm writing it down, but I'm using desmos in place of a calc rn

[tropic oxide]

or better yet, $\frac{a^2+c^2-b^2}{2ac}$

[wraith dagger]

Ann

[thick flint]

And then take the inverse cos of whatever cos(B) is

Yeah
On phone rn tho too lazy to latex

[blissful hawk]

hello guys

i am new here

[tropic oxide]

hello, welcome to the server. you've stumbled into a channel occupied by somebody else rn

if you have your own question then you should take your own channel, see #❓how-to-get-help for how to do that

[blissful hawk]

can i get a bit off topic

[tropic oxide]

#discussion or #chill for off topic.

[blissful hawk]

ok thank you

[potent hazel]

Okay, it would be helpful if someone could help me visualize this triangle

distance and angle, all of it

also when visualing these angles, do i actually need to make it accurate or can I just draw a regular triangle

[tropic oxide]

you can draw a triangle without worrying about scale accuracy

[potent hazel]

im just doing a right triangle trying to identify everything. I do know opposite side and hypotenuse, but it be nice to have someone double check me and place the uppercase letters

[tropic oxide]

this triangle is not right

[potent hazel]

wdym

[tropic oxide]

it isn't a right triangle

none of the angles in it are known to be 90°

[potent hazel]

yes yes i know

Im just saying i drew a triangle

[tropic oxide]

and for the cosine law trying to shove a right triangle into this thing will only confuse you

well anyway if you want to show your drawing then send it here lol

[chrome quail]

Here’s an example with a = 9, b = 8, c = 7

[potent hazel]

My first question bc I'm using desmos is radians or degrees, I think degrees but I want to double check

And then I added and subtracted before getting 2.25 COS (B)

[chrome quail]

order of operations mistake between the last two lines

[potent hazel]

I don't understand how you got a fraction

[chrome quail]

you did 262.44+312.29-573.48

But the 573.48 is attached to cos(B)

so you can't subtract it like that

you need to move the 262 and 312 to the LHS

and then divide out by 573

[potent hazel]

RHS?

[chrome quail]

LHS*

[potent hazel]

I'm sorry I'm still confused

[chrome quail]

left hand side?

[potent hazel]

OHH

Sorry it's late and I already took an exam today 😭

Okay I got 1.3

So I got B= 1.3

[chrome quail]

you did degrees?

[potent hazel]

Yep

I know that that is wrong

[chrome quail]

Then you did something wrong, show your work

[potent hazel]

(16.2^2+13.2^2+17.7^2) / 2(16.2)(17.7) which should equal COS(B)

I see

I didn't subtract b

[tropic oxide]

one of these plus signs in the num should be a minus

[potent hazel]

Im supposed to subtract b^2 which would change the answer to 0.21

[potent hazel]

I just keep getting wrong answer after wrong answer. I'm doing hw, so when I input the answer it just says I'm wrong and to try again to everything

So I made a mistake and the numerator should've been 575.73 instead of 558. I accidentally put 17.2 instead of 17.7 and now it's undefined

[potent hazel]

I put this as the answer as well and it's the same thing to it being incorrect

[tropic oxide]

i didn't take the cos^-1 yet

you can reproduce what i've written in desmos though and get the same number and then also take cos^-1

dont forget to switch to degree mode though

[potent hazel]

Yeah sorry your right, but I showed my work above and I don't see what's wrong it other than the numerator

[tropic oxide]

the 13.2^2 seems to have disappeared here

[potent hazel]

Omg

So it's 45.6

Okay so I'm going to try again with a different problem and I'll post my work and the answer I got for you guys to double check me and whatnot

Where did I fuck up?

[tropic oxide]

,rccw

[tropic oxide]

sign error in line 2

also some fuckery happened with cos(B) with it first disappearing and then reappearing as its evil twin

but line 2 should have been

-2(10.5)(10.8) cos(B) = 12.8^2 - 10.5^2 - 10.8^2

[potent hazel]

Why?

[tropic oxide]

you would be subtracting 10.5^2 as well as 10.8^2 off both sides

at least thats my read on what you were trying to do based on the fact that you left **-**2ac cos(B) as a term on its own

i kind of think you're making your own life way difficult by replacing all of the side lengths with their values immediately and thus burying yourself in a sea of numbers that's very easy to drown in

(and drown in it you do)

[potent hazel]

Yes I'm sure I'm being difficult. I'm sorry I'm really bad at this. I have no idea how I'm even making it this far. Covid and everything fucked up my basic math skills and so I feel like a 7th grader when I'm in fucking college. I digress, but I'll stick to letters for rn and so the first thing I think of doing is moving b^2 to the right and -2ac COS(B) to the left

[tropic oxide]

well, here's what i want you to do then

start with the cosine law formula b^2 = a^2 + c^2 - 2ac cos(B) and then isolate cos(B) in it

while keeping everything as letters

and show me what you get doing so

[potent hazel]

Wait I thought I couldn't do that bc it's part of 2ac?

[tropic oxide]

couldn't do what

[potent hazel]

Like I have to move all of 2ac COS(B)

[tropic oxide]

the isolation will not happen in a single step

the fact that 2ac and cos(B) are multiplied together will eventually be remedied (but not on step 1) by dividing both sides by 2ac.

for now though i will say you should just go and do what i told you to do rather than overthink

[potent hazel]

Okay so it would be divided on both sides

Okay so I got cos(B) = (a^2+c^2-b^2)/2ac

[tropic oxide]

that is correct

and now you can plug your shit in

in fact you may find it somewhat useful to remember that form of the cosine law outright

[potent hazel]

Yes ma'am 🙏🏼

Okay I'm gonna try one more to make sure I got it. thank u sm for the help

Okay so I think i got it, I just got a correct ding so I'm all set

Thanks for all the help! Sorry for the headache but I appreciate the patience

.close

[cedar kiln]

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[worthy forge]

can anyone help me with 9th question

[worldly chasm]

@worthy forge notice that the first term has 2^stuff and the second has 2^stuff but the first stuff is twice as big as the second stuff.

If we call the second stuff y then we have y^2 - 2(a+2)y + 8a < 0

[worthy forge]

oo yeah

then we can do double differentiation

[worldly chasm]

No need, just a regular quadratic. (With y > 0)

You can solve this with 8th grade math

Calculus not required

[worthy forge]

i was trynna solve this yesterday but couldn't solve it

[cedar kiln]

@worthy forge Has your question been resolved?

[cedar kiln]

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[ember storm]

Can anyone help me in 20 min please

[upper ruin]

I would start by multiplying both sides of the equation by the conjugate of z, i.e. $\overline{z}$

[wraith dagger]

Alberto Z.

[upper ruin]

But first you need to state that 0 is a solution, otherwise you can't really multiply as I said

!noans

[cedar kiln]

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.

[lime dust]

It’s a bit late but, I recommend drawing an Armand diagram

Argand

Although I can’t exactly tell you what it’ll look like

I recommend Desmos

Let z=x+yi

[lime dust]

And go from there

[upper ruin]

I believe substitution is the way to go but only after this. Otherwise you have 4th powers

!noans

[cedar kiln]

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.

[upper ruin]

How many times do I have to write it?? @crimson sedge

[crimson sedge]

Sorry everyone

[cedar kiln]

@ember storm Has your question been resolved?

[cedar kiln]

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[glass moth]

I'm new to mechanics

I think you have to set up some equation like:
m*dv/dt = -kv^2-mg ?

because the force acting on the particle would be the net resistive force?

[cedar kiln]

@glass moth Has your question been resolved?

[glass moth]

.close

[cedar kiln]

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[fallen trail]

Can someone please explain to me how it got rid of the proportionality constant, k?

[worldly chasm]

Initial values

Only one value of k satisfies the initial conditions.

[fallen trail]

how so?

[fallen trail]

t=0 so it doesn't make much sense

[worldly chasm]

Oh, sorry I didn't look at your work closely, so I thought your k was the constant of integration. What is it?

[fallen trail]

K is the proportionality constant

[worldly chasm]

The proportionality constant is 4

Not 4k

[fallen trail]

wait how?

when we say x is proportional to 4x what does that mean

x = 4kx right? or x=4x

🤔

[worldly chasm]

It's dx/dt is proportional to 4x

Which, ok so, if I said dx/dt is proportional to x, then that leaves some wiggle room for a proportionality constant, because I wasn't specific

[fallen trail]

yes

[fallen trail]

aaah so then the proportionality constant becomes 4

but is that a good assumption?

[worldly chasm]

If you have 4kx this is no different from kx

[fallen trail]

It's proportional to 4x not equal to 4x

[fallen trail]

yes

[worldly chasm]

Just the identity of k changes

So why give you the 4?

[fallen trail]

yes

[fallen trail]

aaah yeah that makes sense now

But it could've also been 4kx right, although that wouldn't make much sense why it would've said 4

yeah I get your point now

it makes sense

the 4 just becomes useless if we add k

[worldly chasm]

Sure, it's a little ambiguous, it's a reasonable assumption to make

[fallen trail]

yep

tysm

.close

[cedar kiln]

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[cedar kiln]

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[sonic fossil]

For no two alike letters are together to find n(L1 ∩ L2 ∩ L3) is important because there is repetition of letters and to not make many cases we do this

Is my reason correct?

[tropic oxide]

• ❌ missing original question
• ❓ missing definitions of: L1, L2, L3

[sonic fossil]

There is no questions as such, I am using this line to confirm my thinking is correct or not about inclusion-exclusion principle. L1,L2,L3 are some letters for eg A,B,C or O,I,L

[worldly chasm]

So you wanted some intuition about inclusion exclusion?

[sonic fossil]

Yes

[worldly chasm]

The best, most clear explanation of the inclusion exclusion principle I know of is from a larger video by another roof. https://youtu.be/qt5I1gZj1ew 23:40 in

[sonic fossil]

thanks

.close

[cedar kiln]

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[mortal yarrow]

why does X need to be open?

[mortal yarrow]

and also why a function of class C^1 why not use C^infinity to implying that the function is differentiable in all orders?

[dapper raven]

if i remember correctly closed sets have points where functions applied to that set aren't differentiable

[modern estuary]

This is a common assumption at the starting of most multivar calc theorems

In practice even if your function is defined on a closed set, you can restrict to a suitable open set and apply these theorems so the open set assumption is not that much of a problem

[modern estuary]

Because C^1 is enough

[mortal yarrow]

how can that be?

[modern estuary]

Think about the boundary

[mortal yarrow]

but we could easily replace it with C^infinity without affecting the theorem right?

[modern estuary]

Then the theorem becomes weaker, since it applies to fewer functions

[mortal yarrow]

wait so when we use C^1 are we assuming that the theorem applies to all functions that have at least a defined first order partial derivative

right?

[amber parrot]

continuous yes

[mortal yarrow]

ohhh i see whachu mean now

so C^1 is technically the most general case

[amber parrot]

you dont need C^infty

C¹ is enough to make sure the tangent space of S exists at x_0

[mortal yarrow]

yeah this makes sense now

[amber parrot]

X also needs to be open

[mortal yarrow]

i still dont understand why an open set is needed though?

why are functions undefined at the boundary of a set?

[dapper raven]

functions aren't

their derivatives

might be

iirc

[mortal yarrow]

how?

[amber parrot]

because we need S to be smooth locally

[mortal yarrow]

how does a closed set affect its smothness locally?

[amber parrot]

consider S = {x_0}

try proving the theorem is wrong if X is closed

[mortal yarrow]

i see now

thanks

.close

[cedar kiln]

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[rustic plover]

help

[rustic plover]

essentially

i have done part A successfully

i am stuck on the fact on why did he differentiate the function again

since to find turning point,

we just do dy/dx=0

since we found this

would we just equate this to zero

why the need to differentiate it again using product rule

and because they said to find the VALUE and not the NATURE i dont think we need to differentiate it a 2nd time

[modern estuary]

We are finding the maximum value of the gradient(the derivative), not the maximum of the original function. If we were finding the maximum value of the function we'd differentiate just once and set it equal 0

[rustic plover]

whats the difference if i may ask

[modern estuary]

Doing this finds the maximum value of the original function

Thats not what we want to do

The gradient is another function

Whose maximum we want to find

[rustic plover]

so the gradient is a function thats within another function?

[modern estuary]

The difference is that you are finding the maximum value of the wrong function by just differentiating once and making it equal to 0

[modern estuary]

How can one function be inside another function?

I dont understand you question

[rustic plover]

you said here that the gradient is another function so i thought it was inside the original function we found

[modern estuary]

.

[rustic plover]

i assumed

[modern estuary]

They are asking you to find the maximum value of the gradient

The gradient is not the same as the original function

If you just differentiate once and set =0 you get the maximum of the original function

Not the gradient

Which part of this did you not understand, i can expand on it more

[rustic plover]

what im trying to understand is the fact that the gradient is another function and that to find the max/min of gradient we have to differentiate it twice

since thats the 1st time in hearing that

xd

[rustic plover]

differentiate the function twice*

[modern estuary]

We have to differentiate the gradient just once just like we would for finding the min/max of any other function

[rustic plover]

so this is the original function of the gradient?

[modern estuary]

Differentiating the gradient just once is the same as differentiating the original function twice

I am using "original function" to mean the curve which we differentiated to find the min/max of in part (a)

@rustic plover is it clear now?

[rustic plover]

do we only differentiate twice when it comes to gradient and once when it comes to regular functions?

to find max/min

[modern estuary]

Yes

[rustic plover]

so in part (A) it was the first differentiation and in B it was the second differentiation

[modern estuary]

I dont see why this should be a question from you though

Since you know to find the max/min of a function you need to differentiate it once

[rustic plover]

i myself dont know why im so confused at this concept

[modern estuary]

So to find the max/min of the gradient, you just differentiate it once the same way as you do

But differentiating the gradient is the same as differentiating the original function twice

[rustic plover]

the reason im confused is because im thinking that we've already differentiated the function once here

[modern estuary]

Yes

They want you to find the max/min of the derivative

So you have to differentiate the derivative again

Thats another way of saying differentiate twice

[rustic plover]

which is known as f

f''(x)

i know that

[modern estuary]

Yes

they are asking you to find the max/min of f'(x)

So you must set f"(x)=0

Why is this confusing you?

Its the same procedure

[rustic plover]

because we dont set f''(x) = 0 when we want to find max/min, we set f'(x) to zero

we use f''(x) to find the nature

[modern estuary]

We set f'(x) to zero in order TO FIND MAX/MIN OF f(x)

But we want to find max/min of f'(x)

[modern estuary]

So what do we do here?

[rustic plover]

we differentiate it again as that sets f'(x) to zero

[modern estuary]

Ok i will call f'(x) as g(x) from now on

I want to find the max/min of g(x) you agree?

So what should i do

Make g'(x) =0

[rustic plover]

you would find dy/dx of g(x) and equate it to zero

yes

[modern estuary]

But making g'(x)=0 is the same as making f"(x)=0

Thats why the second derivative of f(x) is coming up

[rustic plover]

okay, to find the max/min of GRADIENT we have to take f''(x)=0
and to find max/min of the FUNCTION we have to take f'(x)=0

[modern estuary]

YES

[rustic plover]

well i understand now how to do the questions related to max/min points of gradients but i dont really get the concept behind it

thank you though

[modern estuary]

Its because at a max/min point the gradient of the gradient will be 0

[rustic plover]

yes

OOOOOOOH

[modern estuary]

If your question is solved you can type .solved

[rustic plover]

so essentially dy/dx so that gradient of function can be =0
and f''(x) so that gradient of GRADIENT can be =0

yes?

or is that not it

[modern estuary]

Yes

[rustic plover]

alright tysm

sorry if i was a hassle

[modern estuary]

No worries

[rustic plover]

.solved

[cedar kiln]

Channel closed

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[cedar kiln]

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[ember heath]

🤔?

[dire geode]

new channel, not existing one

[wicked mantle]

it's occupied

[kindred ruin]

yeah his msg was sent while mine was loading, mb

[dire geode]

do you know arctan sum identity

[ember heath]

no i dont think so

i know arc tan x + arc tan y

[dire geode]

i guess try u = sec(x) + cos(x) . then express the other factors in the integrand in terms of u and du

[ember heath]

ok ill let u know what i get

um its not really simplifying

im not able to express the terms in u and du

[dire geode]

why not

[ember heath]

^^

how to convert the sin^3 x?

[dire geode]

find du first

[ember heath]

du=sec x tan x - sin x