help-39

[pearl pond]

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[finite crown]

in my PDE module

[finite crown]

my professor takes the jacobian determinant

but why

we never use it after he takes it

what was the point of the jacobian determinant

am i missing something

[snow sail]

https://tutorial.math.lamar.edu/classes/calciii/changeofvariables.aspx

maybe something like this is helpful

the wiki article on jacobian has a good entry too

[snow sail]

ensuring it is not 0, here

what if it is 0?

[finite crown]

thats what he said but why do we care if its 0 or not?

[snow sail]

LI presumably

im sorry im about to lose service

id suggest #odes-and-pdes

[finite crown]

he said something about it being non singular

[snow sail]

right

try those pages

or ode channel

[finite crown]

will do

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

How do I find the particular integral for this?

[fluid axle]

have you started finding a particular integral and got stuck or no ?

@midnight haven

[midnight haven]

Yh

[fluid axle]

alright where you at then ?

[midnight haven]

I tried 3 times

The first one I tried was, Ctcost + Dtsint + Ecos2t + Fsin2t

Then all the rest were wrong including this one

[fluid axle]

right so what happened with that?

it sounds like an ok guess

[midnight haven]

I differentiated twice then subbed in and compared coefficients, didn't match up with the answer

[fluid axle]

you prolly typo'd somewhere then, it works fine on my part

[midnight haven]

So what I sent is the particular integral

?

[fluid axle]

well what coefficients have you got

that's the whole issue right

[midnight haven]

I didn't finih the question bc I checked the answer and it was going wrong

I'll retry

[fluid axle]

yeah I end up with some coefficients which are 0, so what

[midnight haven]

Do you want to see the answer?

[fluid axle]

sure

[pearl pond]

@midnight haven Has your question been resolved?

[fluid axle]

yeah

that's what I get

using your guess

t cos t - cos2t

[midnight haven]

Can you show me your working plz

I'll send mine

[fluid axle]

we'll work on yours

[midnight haven]

I'm using product rule

@fluid axle

[fluid axle]

yeah there's a few issues there

where do these guys come from ?

[midnight haven]

-2Esin2t and 2Fcost

[fluid axle]

yeah

2E and 2F are just constants right

there's no product rule to make there

[midnight haven]

Ahhhhhhhhhhhhhhhhhhhhhhhhhhhh

Yes I was treating the capitals like variables

Thanks

[fluid axle]

well you weren't treating them as variables for the Cs and Ds

so uhh

[midnight haven]

Ok

Thanks for the help

I'll redo

[fluid axle]

also there's one or two typos here and there be careful

maybe you'll catch them

[midnight haven]

.close

[pearl pond]

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[open rivet]

whats the question?

[woven mantle]

I need to find the convergence domain. Using D'Alembert gives 1

Does raabe duhamell work for finding convergence domain?

So i used that and it worked. Gave me x^2 >1. For x^2=1 the series result in 1/(n+1) which does not converge. Seems right

.close

[pearl pond]

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[pearl pond]

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[leaden wadi]

Probably something stupid like no multiplication betweeen 1/6 and (.

[pearl pond]

@tender cedar Has your question been resolved?

[pearl pond]

@tender cedar Has your question been resolved?

[plush bramble]

Should be bold C

Your i, j, k are bold

[plush bramble]

Show the entire screenshot. Your entry and their answer is cut off

There's a bold button on the right

[pearl pond]

@tender cedar Has your question been resolved?

[dense jasper]

.close

[pearl pond]

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[vivid pilot]

i see how this is true

[vivid pilot]

but idk how to formally prove it

[pearl pond]

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

[vivid pilot]

so i kinda know how it works

[dense jasper]

Think of how you can rewrite $n \equiv 1 \pmod{4}$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

specifically in the form ||n = ak + b for some arbitrary integer k||

[vivid pilot]

n=4k+1

[dense jasper]

Now we want n mod 8, not n mod 4

so what do we eventually want to get to

[vivid pilot]

multiply by 2?

[dense jasper]

kind of

it's a good idea to consider 2

but can you elaborate a bit on that idea

becuase it's a bit vague atm

[vivid pilot]

well im not sure exactly

but like 4k+1 mod 8 is still 4k+1 so its not that helpful

well ig we could multiply the whole thing by smth 1 mod 8

like

9

wait that doesnt help at all eitehr

[dense jasper]

Why don't you consider k modulo something?

Further hint: ||proof by exhaustion||

[vivid pilot]

k mod 8?

[dense jasper]

try it

[vivid pilot]

so

4(1)+1 = 5 (mod 8)

[dense jasper]

?

[vivid pilot]

4(2)+1 = 1 (mod 8)
4(3)+1 = 5 (mod 8)

[dense jasper]

k modulo 8 means that you would consider k = 0 mod 8, k = 1 mod 8, ...

aka k = 8r, 8r+1, 8r+2, ...

[vivid pilot]

4(4)+1 = 1 (mod 8)
4(5)+1 = 5 (mod 8)
4(6)+1 = 1 (mod 8)
4(7)+1 = 5 (mod 8)
4(0)+1 = 1 (mod 8)

so only results are 1 and 5

mod 8

[dense jasper]

uh

read what I said again

[vivid pilot]

oh right

but i mean like

[vivid pilot]

imagine the "=" here was a congruence sign

[dense jasper]

I'm aware of that

[vivid pilot]

then i wouldnt have to do that right

[dense jasper]

My point is that what you're doing is inconsistent with mod 8

ex. if k = 0 mod 8, then you have k=8r, so n=4k+1=32r+1

and hence is n is 1 mod 32, not mod 8

[vivid pilot]

but it would still be 1 mod 8

since 32 = 0 mod 8

[odd surge]

you have n = 4k + 1

you want n mod 8

so you need n = 8j + something

what can you do with k to make this happen

[vivid pilot]

ohh

[vivid pilot]

make k be 2k, 2k+1, so on

4(2k)+1 = 8k + 1

4(2k+1)+1 = 8k + 5

[odd surge]

yep

[vivid pilot]

4(2k + 2) + 1 = 8k + 9 = 8k + 1 (mod 8)?

is that howyou do it

[dense jasper]

why are you doing 2k+2 and onward?

It's not wrong

but there's no practical use

[vivid pilot]

wait why?

[dense jasper]

redundant

2k + 2 = 2(k+1)

[vivid pilot]

hm

but

[vivid pilot]

what about like 2k+3

[dense jasper]

2(k+1)+1

you only need to consider the distinct residues mod 2

aka 0 and 1

[vivid pilot]

oh ok

ty for the help!

.clsoe

.close

[pearl pond]

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[last mortar]

Ok algebra like how do you do it? The question where you are asked if this number is to this what is x? Because that confuses me

[last mortar]

Like x =13 and then they ask u what is x

What is x?

I’m failing math

I have about like a E in math

And I barely answer the questions

[bitter lodge]

x = 13

If youre just given that, you can say x is 13

[last mortar]

x + 3y = 5x what is the x? And how

[hazy pilot]

Jus think of the x as any number

[last mortar]

but what number can it be

[hazy pilot]

imagine 5x = 6 + x, we know that 5 times a number, equals 6 plus that number

[unkempt yacht]

x is the treasure you wish to hunt

to find x you must solve a puzzle; after all, the pattern to the booty is not without difficulty

[fierce totem]

trolling

[unkempt yacht]

the riddler challenges you, what number when adds to 5 results to 8?

[last mortar]

But like how is 5x is 6? Because theirs no number to it cuz for it to be 6 it would be 5+1 is six 😢 I’m so done with algebra

[unkempt yacht]

in your treasure guidebook, it is written has x + 5 = 8

aha! the booty draws near

[jovial zenith]

https://tenor.com/view/black-beard-peak-writing-black-beard-marshal-d-teach-peak-writing-barba-negra-gif-5446515270440960312

[bitter lodge]

Its not algebra, its a ✨ treasure hunt ✨

[last mortar]

but what is x and why do they use those words

[unkempt yacht]

because x sounds cool! you don't hear people say r marks the spot. it's always x marks the spot

of course, the path to the booty involves many such variables

some use y, some use z, but most fundamentally, x marks the spot

[last mortar]

I know but do the variables have a meaning behind it? Cuz I see 5x for example

Is that for decoration?

[unkempt yacht]

well, no

[last mortar]

And why is there a rule behind it where theirs exponents multiplication division addition

😢 bro

[unkempt yacht]

it marks some certain number

for example, can you solve for x? x + 1 = 2

[last mortar]

I don’t even know how to graph on a paper too

[last mortar]

1?

[unkempt yacht]

correct

in this case, x equals to 1. any number it is nonsense

but sometimes x can be changing

[last mortar]

Oh wait so u gotta find the number that equals to that answer? Like the number that actually come into making that answer? Is that what is supposed to be?

[unkempt yacht]

then it is called a variable. it varies

[unkempt yacht]

yeah

2 + 1 = 2 is nonsense, so x is not 2

[last mortar]

Bro is my teacher was this good at explaining I would be acing her damn classes

But I still need help

What is a slope?

[unkempt yacht]

it is a measurement of how steep a graph is

[last mortar]

😞..I failed my quiz I got a zero cuz I didn’t even try

Why do they make a slope like y=mx+b

[unkempt yacht]

because it is useful! thousands of years of maths come to this form, which we found to be the most useful for stuff

[cursive wraith]

To measure the steepness of a line, think of yourself walking along that line
Think about how much effort you need to make to go from left to right

[unkempt yacht]

from this you can very easily see the slope of y is m

[cursive wraith]

if the line is horizontal, you just have to walk straight

[last mortar]

Oh the easiest one would be the no slope one right? 😎

[unkempt yacht]

yes, in that case m = 0

[cursive wraith]

if the line starts shooting up, you might have to do more effort by "climbing" just in order to get the same length to the right

so the way we compute it

is, for some length you want to travel left to right, "Dx"

compute how much extra effort you have to make

by climbing up or down

Call the vertical length you have to travel "Dy"

[last mortar]

I’m getting that a bit but what I need help with most is about the algebra one where they have the exponents and stuff I don’t know what that exponents mean and what to do with them, in my math Nwea test I got a 202 and the average expectations for math was 220 how my world went crushing down after I vowed to make my parents proud 😔

https://tenor.com/view/tony-stark-gif-20710889

[unkempt yacht]

exponent is simply repeated multiplication! you know you can write 5 + 5 + 5 as 5 * 3 right? it is in the same spirit

you can rewrite 5 * 5 * 5 as 5 exponent to 3, usually notated in chat as 5^3

[cursive wraith]

$5\times 5 \times 5 = 5^3$

[jolly parrot]

rafilou is not not born in 2003

[last mortar]

Wait so like if the 5 is at the bottom and a 3 or a 2 is on top of the 2 is on top so I would have to repeat the five 2 times? So 5 times 5? Making 25? Or?

[cursive wraith]

because 5 appears 3 times in the product

[unkempt yacht]

yeah

[cursive wraith]

x * x * x * x... = x^y because x appears y times

[last mortar]

BRO THIS is HOW MY FAM TEACHER OCULD HAVE EXPLAINED IT..was it too much to ask for

Alr so the exponents is how much u will repeat it basically?

[unkempt yacht]

be careful though, exponentiation is not commutative. what does that mean? 5^3 and 3^5 are not the same thing

[hazy pilot]

what kind of teacher do you have 😭

[cursive wraith]

yeah the exponent tells you how many times the base appears in the product

[unkempt yacht]

it is unlike multiplication where 3 * 5 and 5 * 3 are the same thing

[last mortar]

Bro..if I ask a question she’s like “We have been doing this for the past weeks” and she barely CARES

Wait so you know that rule where u have to do division multipcation and addition in order?

Why is that and like is that when u get deeper into it?

Plus the exponents and subtraction

[unkempt yacht]

because we defined it to be so

[last mortar]

Do you love math?

[unkempt yacht]

math is cruel. sometimes she rewards, sometimes she punishes

[hazy pilot]

I love the feeling of solving something

[last mortar]

https://tenor.com/view/himbogravescatwithtearyeyes-himbogravescat-cat-cat-with-teary-eyes-stahp-cat-gif-6378954843936404723

I was never worthy.

I prefer English

[last mortar]

Trust me I don’t 😔

Right now I stay in bed studying

[unkempt yacht]

it is best you do not depreciate yourself

[last mortar]

Plus science is hard for me too

[last mortar]

😔

[unkempt yacht]

you can suck at something, and that's totally ok!

when i was grade 10, i got a 3 in bio and 4.5 at history

out of 10

and it was final exam

[last mortar]

Did you heal?

Have u*

[unkempt yacht]

no. i still suck at bio and history

i'm guessing your problem at math is that you struggle to understand the fundamentals

[last mortar]

Quite the opposites aren’t we?

[last mortar]

I’m sorry but the TEACHER she…she just doesn’t understand me..one time had the audacity to ask if I’m slow made me feel even worthless ..I try I best not to hate her trust me

But that’s what I’ll try to understand.

[unkempt yacht]

sounds like your teacher really sucks. my condolences

[last mortar]

Love the big words. Especially since I’m in advanced enligh class

[hazy pilot]

I used to get only bad grades on my math exams

[last mortar]

That’s what I call pulcritude right there

[unkempt yacht]

what i would suggest is that you try to self study math, especially the concepts you struggle with, when you have free time

[last mortar]

PULCHRITUDE*

[hazy pilot]

until I actually understood what we were doing, which made me more and more interested

Hope everybody gets to experience that feeling sometime in their lives

[last mortar]

But look, when I’m on YouTube searching for math and tips and how to do it, the words come in and leave in my other ear

Promise u.

[unkempt yacht]

try studying at khan academy instead

they have a well-organized course on each concept

[hazy pilot]

I'd look up a visual way of thinking about maths. Number lines and geometry helped me understand algebra better

[unkempt yacht]

and also, this is of utmost importance, pay attention

i'm sure you did, but i just feel the need to clarify

[last mortar]

Words enter my left ear and will leave through my right ear

[hazy pilot]

but pictures won't!

[last mortar]

I’m basically spacing out the entire time

[hazy pilot]

I'd like to teach you right now if you like

tell me anything you don't understand about your class right now

[last mortar]

I tell you I once gotten high motivation and where did that all motivation go? When I went to sleep after studying for one day. Tomorrow I completely forgot that I even was studying abt math

I eat sleep wake up eat sleep

I know different ways of eating 👍

[last mortar]

Please.

🙇‍♂️

[hazy pilot]

What are you talking about in math class right now

[last mortar]

Sorry I was called to go eat

[pearl pond]

@last mortar Has your question been resolved?

[pearl pond]

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[proper flare]

Prove that if a cevian from a vertex of a triangle divides the opposite side in the ratio of the other two sides of the triangle, it bisects the angle of the triangle from which the cevian had been started to be drawn.

[proper flare]

AB/AC = BD/CD = [ABD]/[ADC] = [1/2 (AB AD sin(x))]/[1/2 (AD AC sin(y))]
AB/AC = [1/2 (AB AD sin(x))]/[1/2 (AD AC sin(y))]
Simplifying, we get:
sin(x) = sin(y) = sin(180-x)
Angles of a triangle are in measure between 0 and 180.
sin(x) can be equal to sin(y) in the range 0 - 180 only when either x = y or x = 180 - y [I have concluded this].
x = y is a not abnormal case, so let's look at the x = 180 - y case.
If x = 180 - y, then x + y = 180, which will make angle BAC 180 deg. This implies the other two angles are 0. But side/sin(angle opposite to the side) = 2R
side = 2R * sin(0) = 0
If x + y = 180 deg were to be true, then AC and AB would both have to be 0 in length. The initial ratio 0/0 would be therefore undefined, which is a contradiction to our assumption that the laws of physics are valid. So only the x = y case is correct.
So AD is the angle bisector of angle BAC.
Is this proof correct (of the converse of the angle bisector theorem)?

[unkempt yacht]

what does [ABD] mean

angle?

[proper flare]

[...] area function.

[unkempt yacht]

okay

[odd surge]

laws of physics is crazy

[proper flare]

<@&286206848099549185>

You know what's more crazy? Someone actually helping me. (roasted)

'Laws of Physics' was just a joke if anyone got confused or thought it involved Phyiscs (Physics aversion, roasted again)

<@&286206848099549185>

I need help with this question:
Is my proof of the converse of the angle bisector theorem complete and is everything justified in the proof which needs to be?

[pearl pond]

@proper flare Has your question been resolved?

[proper flare]

This question isn't very hard, is it?

[pearl pond]

@proper flare Has your question been resolved?

[pearl pond]

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[pearl pond]

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[minor adder]

<@&286206848099549185>

not sure what "find two equations involving alpha" means

[toxic lichen]

do you know vieta's formulas?

[minor adder]

no

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[minor adder]

soneone help pls

[timber cosmos]

ping a helper in 5 min

[minor adder]

can u help

[timber cosmos]

uh sure

gimme asec

[toxic lichen]

sum and product of roots? no?

[minor adder]

hmmm

[timber cosmos]

you know vietas formulas

?

[minor adder]

so like sum divided by 3

[toxic lichen]

he already said no

[timber cosmos]

oh

[toxic lichen]

you're trying to jump ahead

let's step back

[minor adder]

ok

[toxic lichen]

for the equation ax^2+bx+c=0

what is the sum of its two roots equal to

[minor adder]

-ba

-b/a

[timber cosmos]

whats the prodcut

[toxic lichen]

yes, -b/a

and their product?

[minor adder]

c/a

[toxic lichen]

these ARE vieta's formulas, just so you know

[minor adder]

oh

ok

[toxic lichen]

write them out in the case of your question

and take the roots as alpha and 2alpha as it says to do

[timber cosmos]

i got 3a=6/a

for the sum of the roots

what how

[toxic lichen]

do not confuse a with alpha

[minor adder]

wait so -2/a = 2alpha^2

[toxic lichen]

no

[minor adder]

and 6/a=3alpha

[toxic lichen]

(a-2)/a not just -2/a

[toxic lichen]

this one is correct

[minor adder]

oh yeah

[timber cosmos]

thats what i did

[toxic lichen]

no you confused the letters a and alpha

[timber cosmos]

after this can you help me in my channel

2 below this

after this

[minor adder]

ok so a-2/a = 2alpha^2

[toxic lichen]

!noadvert

[pearl pond]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[minor adder]

and 6/a=3alpha

[toxic lichen]

missing parentheses here

[minor adder]

whats that

[toxic lichen]

() these things

(a-2)/a

[minor adder]

brackets

[toxic lichen]

otherwise it reads like $a-\frac 2a$ and you don't want that

[jolly parrot]

ann.in.a.teacup

[minor adder]

ok

[toxic lichen]

anyway yeah you got those two equations

so part a) is done

[minor adder]

ohhhh

that what it means

do i use simuletanoues equations or smth

[toxic lichen]

"use simultaneous equations" is a silly turn of phrase

a set of simultaneous equations is not something you really "use". it always arises naturally by itself.

but yes i suppose that you do.

[minor adder]

ok

is there any othr way

👍 good chat

wait

i got a =4 or -2

is that right

how is there 2 a's

[toxic lichen]

show your work?

preferably on paper

[toxic lichen]

is there any reason why you chose not to simplify 6/(3a) into 2/a

[minor adder]

uh

i forgor

but that wouldn't really change it would it

[toxic lichen]

anyway yes your two values of a are correct anyway

[minor adder]

how is there 2 values of a

wont there be 4 values roots then

[toxic lichen]

there will be two pairs of roots

[minor adder]

oh dang

[toxic lichen]

which are equal to 2/a and 4/a

[minor adder]

okay i think iknow what to do now

[toxic lichen]

we can look at the original equation and solve it in both cases (ie for a=4 and for a=-2) if you want

see if there's something we both missed that makes it NOT satisfy the "one root is double the other" thing

in either case

[minor adder]

nah i dont miss stuff

谢谢

.close

[pearl pond]

Channel closed

Closed by @minor adder

Use .reopen if this was a mistake.

[toxic lichen]

you're welcome

[minor adder]

chat i tihnk i missed something

[pearl pond]

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[dapper kraken]

find all positive integers n such that n^2+3^n is a square

[dapper kraken]

m^2=n^2+3^n
m^2=n^2 mod 3, 0=0 mod 3, 1=1 mod 3, 2=1 mod 3, so either both m and n arent multiples of 3, or both are

m^2=n^2+(-1)^n mod 4, i dont think this one helps tho

none of these really helped, another idea was (m-n)(m+n)=3^n where m-n=3^a and m+n=3^b, a<b and a+b=n, n=3 works

[sharp smelt]

$n^2+3^n=m^2$
\
$(n-m)(m+n)=(2+1)^n$

[jolly parrot]

What a wonderful world it is !

[sharp smelt]

Then maybe binomial the RHS

[dapper kraken]

wtf?

[sharp smelt]

nah, that won't work

[dapper kraken]

i think you can factor 3's from both n and m and cancel them with the 3^n so that gcd(3,n) and gcd(3,m)=1 (although this kinda loosens the 3^n so im doubting if it helps)

[pearl pond]

@dapper kraken Has your question been resolved?

[dapper kraken]

.close sorry gtg

[pearl pond]

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[pearl pond]

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[proper flare]

?

[proper flare]

I was giving an exam. I am back.

[proper flare]

Please refer to this old thread.

It's just a proof of the converse of the https://en.wikipedia.org/wiki/Angle_bisector_theorem.

Please see this also, though the priority is of the proof of the angle bisector theorem (and its converse), since it's used in proving the incircle theorem I had proved yesterday.

[pearl pond]

@proper flare Has your question been resolved?

[proper flare]

<@&286206848099549185>

[pearl pond]

@proper flare Has your question been resolved?

[proper flare]

<@&286206848099549185>

[pearl pond]

@proper flare Has your question been resolved?

[plush beacon]

tengo una duda con razones trigonometricas

[compact ridge]

(answered)

[pearl pond]

@proper flare Has your question been resolved?

[pearl pond]

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[signal peak]

i dont know if this is a stupid question but could you argue that \sqrt{\left(-1\right)^2} could equal -1 and one depending on how you simplify it

[pine nebula]

$\sqrt{\left(-1\right)^2}$

[jolly parrot]

blindInjection

[light helm]

no

[signal peak]

what if you just cancel 2 with the square root

doesnt that leave you with -1

[light helm]

that rule only applies if the base is non-negative

[uneven kindle]

A root and a square dont cancel eachother

A root always needs a nonnegative

Taking the root of a square makes what was squared absolute

[signal peak]

oh

thanks

.close

[pearl pond]

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[rare anchor]

what would be the general solution to the functional equation f'(f(x))=f'(x)? i know ax+b is one of the possible solutions but i was wondering if there were more to it

[rare anchor]

f is differentiable at all points

oh and uh let's also say that f'(x) is continuous as well because for all real number k f(x)=ax+b (x!=k) and ak+b(x=k) would also be a solution without the aforementioned condition

[prisma elk]

any conditions on f? @rare anchor

[rare anchor]

as long as f is differentiable at all points and the derivative is continous, not really

[prisma elk]

well both f(x) = ax+b and f(x) = c are solutions to it @rare anchor

[cinder bane]

i think this seams like a differential equation

[rare anchor]

yeah that's like what i could get, was wondering if there were more to it. can't find any more than that 😔

[cinder bane]

I still don't see how can it be solved but maybe this might help also
$$\frac{d}{dx} f(f(x)) = f'(f(x)) f'(x)$$

[jolly parrot]

Sherif Player

[cinder bane]

So mixing that with the original equation we get
$$\frac{d}{dx} f(f(x)) = (\frac{d}{dx} f(x))^2$$

[jolly parrot]

Sherif Player

[rare anchor]

oh interesting, i think you meant $$\frac{d}{dx} f(f(x))=(\frac{d}{dx} f'(x))^2$$ but thanks for the insight

[jolly parrot]

skullemoji

[prisma elk]

no he meant what he said earlier

[cinder bane]

nah
it was
$$f'(f(x)) * f'(x) = f'(x) * f'(x)$$
subistitute the first equation i got
$$\frac{d}{dx} (f(f(x))) = (f'(x))^2$$
which can be writen as
$$\frac{d}{dx} (f(f(x))) = (\frac{d}{dx}f(x))^2$$

[jolly parrot]

Sherif Player

[rare anchor]

oh right im stupid

it's 2am forgive my dumb ahh

i didn't see the dx in front of it

[cinder bane]

now i wonder if there is a way to cancel these d/dx

[rare anchor]

generally given any function g(x) and h(x) if g'(x)=h'(x) then g(x)=h(x)+C

oh the dx is inside

nvm

[cinder bane]

but here g'(x) = (h'(x))^2

[rare anchor]

yeah... that's tricky

[cinder bane]

is there a way to solve that with differentail equations?

[rare anchor]

maybe, correct me if i am wrong because i am just a high school student, i don't think there is a conventional way of integrating f(x)^2 in this scenario

[cinder bane]

maybe we can use laplace transformation

[prisma elk]

Thats overkill kekw

[rare anchor]

most of the time differential equations don't have composite functions so that's where it puzzles me

[prisma elk]

there cant be more, try supposing there is non-linear solutions & find a contradiction

[rare anchor]

that's an interesting perspective

i'll try that out, thank you both of you for the help!!

.close

[pearl pond]

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[prisma elk]

yeah suppose f'(x) is non constant, work ur way out

[pearl pond]

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[left idol]

Integral called i

i gotta justify, without calculating it, that i is between 0 and 1/2

deadass i've gone over exercises liek this a couple times and i still find em hard

[toxic lichen]

one side of the inequality is much easier than the other

[left idol]

ye

saying that is higher than 0 is way easier

because the root has to be 0 or higher

etc etc

[toxic lichen]

as does x

yes

[left idol]

how about the 1/2 tho

[toxic lichen]

well, here is a trick

$ab \leq \frac{a^2+b^2}{2}$ for any real $a$ and $b$

[jolly parrot]

ann.in.a.teacup

[toxic lichen]

try to use that with some clever choice of a and b, and see what happens

[left idol]

⁉️

[toxic lichen]

(this is the inequality 0 ≤ a^2 - 2ab + b^2 rewritten)

(and a^2 - 2ab + b^2 = (a-b)^2, which is clearly always ≥ 0 as a square)

[left idol]

yeah im lost

idk

[green dew]

u can try analyzing the integrand

and see its monotony in different intervals

[left idol]

honestly i didnt wanna do the monotony tho

but yeah that works

[green dew]

i tried it

[green dew]

soz ig 😔

[left idol]

i searched up and found that solution but

i wanted to know what ppl had to say here

like if there was an easier alternative or sm

[green dew]

probably

there is

but i js do traditional french resolution of problems

[left idol]

i dont get it tbh @toxic lichen

[prisma elk]

$\forall x \in [0,1] \sqrt{1-x^2} \leq 1$

[toxic lichen]

i dont really know what to suggest without the trick

you need x^2 @prisma elk

[jolly parrot]

Goëtia

[prisma elk]

mhm?

u multiply by x, u get the integral of x from 0 to 1

it is equal to 1/2

done

na?

[toxic lichen]

it says to justify WITHOUT calculation

thats the thing

oh

uhhh

ok youre saying to majorize just the root by 1 and THEN integrate?

hm

[prisma elk]

yes

well let me think of something less "numerical" XD

[toxic lichen]

yeah i guess that works but it kinda walks the line

[left idol]

by solving they mean like to actually sove the integral ig

so that would be fine no

?

[prisma elk]

i have an idea

consider f(x) = x*sqrt(1-x²)

when is this function reach its max? @left idol

[left idol]

thats the same as monotony

[left idol]

but yae

i might do it taht way

[prisma elk]

wym monotony? @left idol

[left idol]

read from here

[prisma elk]

u can use what Ann-san suggested

you know AM-GM? @left idol

[left idol]

no

[prisma elk]

ok suppose a,b > 0

then (a-b)² >= 0 right ? @left idol

then a² + b² >= 2ab <=> (a²+b²)/2 >= ab, then now put a = x, b = sqrt(1-x²)

[pearl pond]

@left idol Has your question been resolved?

[pearl pond]

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[midnight haven]

hi

[midnight haven]

give the negation of this statement

i got

there exist (x, y) such that xy = 0 or x more than or equal to y implies 1/y > 1/x

is this right

yes

[spare lark]

Well, the negation of P => Q is P and not Q

[midnight haven]

oh

good point

[spare lark]

As a hint There should have no implies in the answer

[keen rain]

note A => B is the same as -A OR B

[midnight haven]

soo...

xy ≠ 0 and x < y implies 1/y > 1/x?

[keen rain]

hm why with implies

[midnight haven]

ohhhh i get what you mean now

i should say and instead of implies right?

-(-A or B) = (A and -B)

[keen rain]

yup

[midnight haven]

great

thanks

.close

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[cobalt tide]

Hi I just learned about perpendicular and parallel vectors and was wondering something.

So if n and m are perpendicular
and (direction vector)m = (normal vector)n

Does that mean that (normal vector)m = (direction vector)n?
Would you be able to solve it that way too?

[pearl pond]

@cobalt tide Has your question been resolved?

[cobalt tide]

.close

[pearl pond]

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[scarlet radish]

how do I know how to guess a and b?

[open rivet]

use b0 and b1

b0=a0+b=1

b1=a1+a+b=1

[scarlet radish]

yeah i got that. but surely theres infinite solns to those equations so how can i fix alpha beta

[open rivet]

is a_n given?

[pearl pond]

@scarlet radish Has your question been resolved?

[rancid grail]

usually you do that by using b0 and b1

honestly though i would just find zeroes of the characteristic polynomial and solve it that way

[open rivet]

!noadvert

[pearl pond]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

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[normal sigil]

Hiya. I want to prove this statement (but here I'm just trying to show that the left-hand limit exists first). I have no doubt that my proof for this is verbose and might contain unnecessary steps. Is it correct?

[pearl pond]

@normal sigil Has your question been resolved?

[normal sigil]

<@&286206848099549185> Sorry. Apparently I can ping you after 15 minutes.

[toxic fractal]

on second line, what's x_0 supposed to be?

just any point in (a,b)?

[normal sigil]

Oh, well I'm trying to show the left handed limit as x -> x_0 exists for any arbitrary x_0 in (a,b).

So I guess so.

[toxic fractal]

You need to specify so. Also, i think you mixed those two up

[normal sigil]

Oopsie, yes

Think I got it right in the body of the proof though

[sudden heath]

other than that it looks correct

[normal sigil]

Is the business with the subsequence necessary?

[sudden heath]

dont think so

but its valid

[normal sigil]

I had the subsequence and I wanted to have the actual f(a_n), so I just did this. And now I don't think it even is necessary

[toxic fractal]

i mean, you explicitly asked if it was correct, regardless of verboseness

[normal sigil]

Well, sure. Then I asked a follow-up question: 'Is the business with the subsequence necessary?'

wrong emoji

[toxic fractal]

fair enough

[sudden heath]

well yes if you argument with sequences

[normal sigil]

Anyway, I think I can just use f(a_n) directly right

oh

[toxic fractal]

you also need to specify what a_n is

[normal sigil]

Well, it's a sequence in [a, x_0) with a_n -> x_0.

I thought I could just write that

Is there something I forgot to specify?

[toxic fractal]

you need to specify that a_n is a sequence before saying that it has a subsequence

[normal sigil]

Oh, I suppose you're right. Maybe I elided a few words there

[sudden heath]

i mean an -> x0 really already implies it being a convergent sequence

[normal sigil]

I might as well show you this. Here's the hint: I wasn't sure how to extend it from the 1/n one to all sequences, so I went my own direction

I am not sure what the hint was trying to get me to do

Any idea how I would 'show that any other sequence' satisfies the same thing?

[toxic fractal]

it can be a function, or a number

[sudden heath]

its pretty clear i think its okay

[toxic fractal]

also, are these strict inequalities?

[sudden heath]

later in maths you leave out half the jargon anyway if its clear whats meant

[normal sigil]

I should hope I won't get marks taken off for such omissions in exams...

[toxic fractal]

on exams it completely depends on what explicitly you were asked before, and how rigorously you're expected to prove it

[normal sigil]

Well, we don't consider a_n = x_0 when we are looking at the limit, and I think if the left < were \leq instead it'd really be the same thing.

[normal sigil]

That's true. I think it'd probably be acceptable to rely upon the context that 'a_n is definitely a sequence' but I guess I should make it explicit in case.

[toxic fractal]

well, if you're saying a_n converges to x_0, the limit should be able to be equal and thus the inequalities should not be strict, maybe?

[sudden heath]

you might want to point out why there is an increasing subsequence tho

[toxic fractal]

a completely rigorous proof needs everything to be explicit

[normal sigil]

In the definition of limit we have 0 < |x - a| < \delta, explicitly getting rid of the consideration of x = a. I therefore thought that in this case I would have this particular inequality strict to match up with this sort of thing.

[normal sigil]

Yes, makes sense. I will add that in.

[normal sigil]

Should it be non-strict?

[sudden heath]

now that i look at it, your proof that f(an) converges looks a bit weird

[normal sigil]

Hmm... if I have the x_0 - \epsilon < a_n < x_0, maybe I should write f(x_0) - \epsilon_2 < f(a_n) < f(x_0) instead?

[toxic fractal]

the definition of limit does not include the 0<

[sudden heath]

it does

limit doesnt require the function to be defined at point a

[toxic fractal]

no, but it doesnt require the absolute value of the difference to be non-zero

otherwise, functions like the constant function would not have a limit

[normal sigil]

They would because it's x not f(x)

[sudden heath]

the absolute value is in the domain

[normal sigil]

This is what the textbook I'm using says:

So the 0< only makes one appearance.

[toxic fractal]

that's not for the limit, that's for the value you're approaching

[sudden heath]

thats what he said

[normal sigil]

I'm a bit confused.

[toxic fractal]

that was not what i was talking about tho

[normal sigil]

Maybe my proof structure is what is confusing.

[sudden heath]

i would suggest you start with an abitrary epsilon and show that at somepoint f(an)-l < epsilon

[toxic fractal]

the 0< is included on the "what you're approaching" to exclude being at the point. aka, the function/sequence doesnt need to be defined at the point
It's not included in the limit itself, since you can get close enough to be exactly on the value to be approached

[sudden heath]

but where does he claim that

[toxic fractal]

we were talking about different things

[normal sigil]

Well my usage of the strict inequality is in the 'approaching' part, which is this.

And then when I apply f, is that the problem that it doesn't become non-strict?

[sudden heath]

no it stays that way

the problem is

you just gave an upper bound for f(an)

that doesnt mean it converges

[normal sigil]

But I can write f(a_n - epsilon) = f(a_n) - \epsilon_2 and if epsilon decreases then epsilon 2 decreases. Although... my proof is seeming very contrived now.

[toxic fractal]

didnt he tho? he also gave a lower bound and said for all epsilon, which means arbitrarily small

[sudden heath]

he didnt say why f(x0-epsilon) as a lower bound makes it arbitrary small

basically he asumed what hes trying to prove

[normal sigil]

If I did explain why f(x0 - epsilon) does make the range of allowed values for f(a_n) narrower would this fix the proof?

[sudden heath]

hmm i think it would be a bit complicated that way

[normal sigil]

I will try to start again using this hint.

[sudden heath]

your proof is fine up until that point

[normal sigil]

I only did that because I couldn't think of an explicit way to link a_n converging to f(a_n), but maybe I shouldn't really use sequences for this

Thanks.

.close

[pearl pond]

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[sudden heath]

hey hold up

[normal sigil]

I'm back!

Do you want me to reopen it?

[sudden heath]

yeah

[normal sigil]

.reopen

[pearl pond]

✅

[sudden heath]

so you showed, that there is a subsequence of f(an) that converges to l

[normal sigil]

Yep.

I did that because all the subsequences have to converge to the same limit if the original sequence converges

[sudden heath]

you know what i think there is a much better way

[sudden heath]

just argument, that an is increasing except for finitely many n and thus is f(an) -> f(an) converges because its increasing subsequence which differs by finitely many elements converges

[normal sigil]

How do we know that a_n is increasing except for finitely many n? Couldn't it double back by a small amount and then get even closer to l?

[sudden heath]

oops nope youre right

[normal sigil]

I did think that it would be 'eventually monotone' at first but then I discarded the possibility.

[sudden heath]

its a little late for me already 😅

[normal sigil]

Analysis after 11 pm at the latest is bad for the brain

That's my rule.

[sudden heath]

its 12 pm for me oops

[normal sigil]

Dearie me. Your energy levels must be converging to 0

I'll close this again. Thanks for your help

.close

[pearl pond]

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[unreal basalt]

I don’t get this is this congruent or supplementary

[unreal basalt]

Idk how to write the equation

[open rivet]

you have 180=X+O

[unreal basalt]

What is X and O

[open rivet]

look at the drawing imade

"drawing"

[unreal basalt]

I don’t understand it

[open rivet]

what part?

its F-angles

[unreal basalt]

Is the equation 180=5x - 24 + 89

[open rivet]

yup

then solve for x

[unreal basalt]

Ok I’ll just solve it from here ty

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[quartz yoke]

could someone please help me on this

so i made the logx x^2 to just 2

and then i did the change of base formula and aded them together

am I on the right track?

[rancid depot]

added?

[quartz yoke]

yep

[rancid depot]

can you show your work?

[quartz yoke]

sorry ignore that i substituted everything for sqrt71

i shouldnt have done that

[rancid depot]

ah

you confused log(ab)=log(a)+log(b) with log(a)log(b)=log(a)+log(b)

[quartz yoke]

crap

[rancid depot]

if you keep everything in parentheses, the change of base will give you fractions that cancel out

[quartz yoke]

omg i see that

thanks im gonna do that real quick

[rancid depot]

in fact, youll notice that x can be whatever you want

y will still be the same

[quartz yoke]

nice i got it

thanks a ton

i appreciate it

[rancid depot]

y=?

[quartz yoke]

9

[rancid depot]

nice

youre very welcome

[quartz yoke]

thanks

.close

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[pearl pond]

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[pearl pond]

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[kind barn]

Need serious halp

[kind barn]

This is the question, very stuck

Answer is bottom left, 706.913

Hold up, I think I figured it out

[proper flare]

<@&286206848099549185>

[rain vessel]

Wait, brother, why don’t you open a new help channel😭

[rain vessel]

If you still need help you can keep this channel open, but if you no longer need help please close the channel

[proper flare]

There's a command for saying this:
!done

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

@kind barn Has your question been resolved?

[kind barn]

Not done sorry

Tried new solution

Also wrong

Still need help

[rain vessel]

Can you post the og problem?

[kind barn]

Orrr..

better quality

I really dunno wot to do at this point

I feel like I've tried everything I know so

[odd surge]

sine rule

[kind barn]

"Let ABC be any triangle with a, b and c representing the measures of the sides opposite the angles with measures A, B, and C respectively. Then, the following is tru:

SinA/a = SinB/b = SinC/c

I keep doing it and get 706.91 or 706.24

Can someone tell me maybe where im messing up?

[kind barn]

dis mah work

[kind barn]

Can you give me like

A super indebth describing of whatever the sine rule even is?

<@&286206848099549185>

Frick it

Photomath

[simple tide]

if you can find a 'r' value then you will find out ab distance

[kind barn]

Still confused sorry

[simple tide]

or use cosine rule

[kind barn]

I still don't get it

Ima just get help tomorrow from the teach

[kind barn]

Thank you

.close

[pearl pond]

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[dapper kraken]

status 1

found this online and looked fun

[vital crescent]

whats status 1

[dapper kraken]

!status

[pearl pond]

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

[vital crescent]

ig a first thing to check is what are the conditions you can check for concyclicness. there are a few on angles and a few on length. do you have a hunch on which one might be easier to do?

[dapper kraken]

icl none of them look easy cause the angle of the midpoint of a line is abit tricky to find no?

[vital crescent]

well so maybe try length first

[dapper kraken]

err length?

the only one i know is ptolemy 😭

[vital crescent]

like FEBC concyclic because FH *HC=EH*HB

[dapper kraken]

oh poap?

[vital crescent]

😭 idk names/abbriviations but probably

[dapper kraken]

power of a point

[vital crescent]

yeah i think so

I don't remember much nouns from geometry... I remember them as some facts

[vital crescent]

so if you do it on AHMD, at the very least, there is less to worry about for "angles"

[dapper kraken]

can you even apply that here at AMHD

[vital crescent]

connect AH? and let it intersect DM?

[dapper kraken]

yeah but like eww

im not sure how to find any of the lengths

call the intersection N

how can you find AN, NH, MN, DN

[vital crescent]

you want AN*NH=DN*NM, well you can translate AN*NH, you should try a bit see what you get

[dapper kraken]

translate?

[vital crescent]

i mean there is some equality you can make with AN*NH using cyclic quad

try around see what you get. If you are really stuck I can give you more hints

[dapper kraken]

equality?

[vital crescent]

yeah

for example, you can make an equality with FH*HC: FH*HC=EH*HB

[dapper kraken]

FN×NE=AN×NH (this seems promising?)

yea idkr

i dont see anything cocyclic that has EF and uses N

i dont see anything with DM also

WAIT

N is on AH, which is the radical axis of (AMHD) and (AFHE), so power of N on both (AMHD) and (AFHE) is the same

the power of N to AFHE is NH×NE, and the power of N to AMHD is ND×NM, and since the power is equal, ND×NM=NH×NE=AN×NH

[vital crescent]

NH*NA rather than NH*NE also you do not have AMHD concyclic

thats the thing you are trying to prove

[dapper kraken]

ugh right

cant i say (ADH) instead

[pearl pond]

@dapper kraken Has your question been resolved?

[vital crescent]

sry i was away for a bit

what do you mean by that?

here is a graph

We want to show that DN*NM= AN*NH.
But AN*NH is just FN*NE
So our goal is to show that DN*NM=FN*NE and we will be done.

I wanted you to play around for a bit with this expression and see what you get. If you want a hint: ||you can get DN*DM=DF*DE||

[dapper kraken]

lemme try to reword it

[twin sigil]

Try

But do not stata

[vital crescent]

stata?

[dapper kraken]

AN×AH=FN×NE, since for circle AHE and AHD the radical axis is AH, and N is on the radical axis, then AN×NH=DN×MN

ok i dont think this works

[dapper kraken]

how did you get DN×DM=DF×DE?

[vital crescent]

ANNH=DNMN builds on the assumption that AMHD is cyclic

[dapper kraken]

yeah mb

[vital crescent]

try playing around for a bit, see if you can get it first

usually at this stage you should try and "brute force" and find some nice equalities, which is what I did

[dapper kraken]

DF×DE makes me think power of D to (AFE)

i want some sort of circle that contains M and N such that the radical axis of (AFE) and it is FE, but i dont think one exists

[vital crescent]

you should really do it on cyclic quads, AFE doesn't seems to lie in a nice cyclic quad

[dapper kraken]

AFHE no?

[vital crescent]

true, but, there this doesn't help in changing DF×DE into something else

On the other hand, the power of D on (FECB) is a lot nicer

you immediately have DB*DC= DF*DE

but let us worry about that later.

[vital crescent]

try and get DN*DM=DF*DE, If you are stuck lmk

and your intuition is correct, this form is "nice" because it looks like both LHS and RHS are power of D.

[dapper kraken]

going backwards from end result BNMC is cyclic, but i genuenly have no idea how i can prove that

[vital crescent]

yeah but let's do it one step at a time (there are other cyclic quadrilaterals that are nicer)

[dapper kraken]

thats the only cyclic quad that seems useful to me ;-;

[vital crescent]

have you proven that DN*NM=FN*NE happens if and only if DN*DM=DF*DE?

[vital crescent]

don't worry about it for now

I will try and guide you without revealing too much

[dapper kraken]

wtf 😭

[vital crescent]

is some bit confusing?

[vital crescent]

or do you want hint to show this?

[dapper kraken]

a hint would be nice

i cant think of a way to even try to tackle that

[midnight haven]

Hi

May I help?

Whats the question

[dapper kraken]

this

[midnight haven]

Dang proof
I am sorry I suck at that

[vital crescent]

usually it might be nicer to put it into a fraction. say, DN/FN=NE/NM so this is equivalent to DF/FN=ME/NM,

and this is equivalent to DF/(FN+DF)=ME/(NM+ME) etc

[dapper kraken]

how is DN/FN=NE/NM equivalent to DF/FN=ME/NM what

[vital crescent]

(DN+FN)/FN=(NE+NM)/NM

really think of these as ratios

[dapper kraken]

AA_Sus

[vital crescent]

proving a:b=c:d is equivalent to proving a+mb:b=c+md:d

😂 maybe this is too hard to understand, if so I can give you another method

[dapper kraken]

shouldnt it be -

[vital crescent]

nice catch mb

yeah

(DN-FN)/FN=(NE-NM)/NM

[dapper kraken]

which is equivalent to DF/DN=ME/NE...

idt you can repeat the trick again?

[vital crescent]

\begin{align*}
DN\times NM= FN\times NE&\iff DN:FN=NE:NM\
&\iff DF:FN=ME:NM\
&\iff DF:DN=ME:NE\
&\iff DF:ME=DN:NE \quad (Why?)
\end{align*}

[jolly parrot]

qwertytrewq

[vital crescent]

you can also swap the ratio like I did in the last line, hopefully you can see why

this is very close to $DF:DM=DN:DE$ which is what we wanted

[jolly parrot]

qwertytrewq

[dapper kraken]

you can do the trick to get DF:(DF+ME)=DN:DE, and by definition ME=MF so DF:DM=DN:DE

[vital crescent]

perfect!

(btw you should be able to prove why these "tricks" works)

So our attention is turned to proving DN*DM=DF*DE

it would be nice if NM and two other point X,Y are concyclic, where DNM, DXY are collinear (so we have DN*DM=DX*DY)

and if magically FEXY are also concyclic, then we have DX*DY=DF*DE, in which case we will be done!

So the issue we currently have is finding a pair of X,Y that are nice.

[dapper kraken]

B and C?

[vital crescent]

One candidate is setting X=B, Y=C, but it seems hard to prove that BNMC are concyclic.

[dapper kraken]

ok

[vital crescent]

what other circle do you know that pass through F and E?

[dapper kraken]

AFHE... but this breaks ANHM

[vital crescent]

yeah we really want the other two points to be collinear with D

otherwise we don't have DX*DY=DF*DE

maybe also think about what X,Y will be nice with N,M

one natural candidate is to set X=I since NI is perpendicular to DI (right angle has to be nice, right? lol)

so all you need to do is find Y on the ray DX such that YM is perpendicularto DM

[dapper kraken]

is it G

[vital crescent]

you gotta be more specific lol i never told you what G is

[dapper kraken]

midpoint of BC

[vital crescent]

but yeah it is G

[vital crescent]

yeah!

do you know why GM is perpendicular to DM?

[dapper kraken]

no :D

[vital crescent]

hint: G is the center of the circle (FBEC).

[dapper kraken]

oh

M is midpoint of chord EF and midpoint of chord to center is always perpendicular to chord

[vital crescent]

yeah!

anotherway to know for sure Y has to be G is to notice that FEIG is concyclic by 9 point circle

so there gotta be a way to show NMIG is concyclic

[dapper kraken]

how do you prove FEGI is concyclic

[vital crescent]

cuz it is part of the 9 point circle?

[dapper kraken]

time to google!

[vital crescent]

oh you didn't knew about it?

[dapper kraken]

no

ok

so NMGI and FEGI is concyclic

[vital crescent]

it more or less just showing that the midpoint H and centroid is the center of the 9 point circle

[dapper kraken]

DN×DM=DI×DG=DF×DE

[vital crescent]

yeah and backtracking everything you are done

[dapper kraken]

indeed

[vital crescent]

afaik you just let O to be midpoint of H and the centroid, and prove O to the 9 points have the same length

a pretty boring proof irrc

[dapper kraken]

ok

ty <3

.close

[pearl pond]

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