help-39

[silk vine]

can i see

[mighty meteor]

here

[silk vine]

thing is

you got the answer right

using the substitution earlier

[mighty meteor]

this too

[mighty meteor]

i for sure, 100% didnt

[silk vine]

this is right taking into account

its all multiplied by 2

and +1

instead of -1

[mighty meteor]

oh right

oh wait

thats right

my bad

[silk vine]

its not the form they want

[mighty meteor]

would this work tho?

[silk vine]

but i think its pretty good

[mighty meteor]

like, i dont wanna redo it for nothin

[silk vine]

im pretty sure your other solution is right too

ill have a look

[mighty meteor]

ill just redo it then

[silk vine]

the whole thing looks right

im pretty sure its the same form but I wouldn't know how to show the 2 are the same using trig identities

[mighty meteor]

ill show u this time

btw, correct answer is d)

so like

i think i got it

[silk vine]

aw man

[mighty meteor]

ill show u

[silk vine]

appreciate it

[mighty meteor]

here you go

ignore the first line

its wrong

it starts from the second line

[silk vine]

thats very nice

[mighty meteor]

yup

it is

ill send you the next one

if you get tired

or bored

or just dont wanna do anymore

[iron stream]

Yo hii new question?

[silk vine]

nah i dont mind

[mighty meteor]

then just tell me

[mighty meteor]

yea, in a bit

[silk vine]

I need to prep for the integration bee antway

[mighty meteor]

this ones should be way easier, thats what the book thinks at least

ill send them now

[silk vine]

so as much integgrtion as possible i good

[mighty meteor]

i could send you these hell exercises from this book

oh shit

my phone is dying, imma go grab a charger

brb

[silk vine]

i have enough hell exercises

but ill take some more

[mighty meteor]

im on my laptop, but like i need it to take pics

[mighty meteor]

makes sense

also, i dont understand volumes with integrals, but ill leave those for tomorrow

[silk vine]

volume integrals are actually not to bad

cause its basically nothing new

cause they wont make you do a difficult substitution

[mighty meteor]

here the exercise

[mighty meteor]

yea, i just dont think we do them in school

but they require it for the uni exam

which means that i gotta study them by myself

[silk vine]

yikes

[silk vine]

doesnt seem too bad

[mighty meteor]

it shouldnt be

[silk vine]

splitting it up seems to work nicely

[mighty meteor]

ight

imma do exactly that

i think i know what ure talking about

[iron stream]

The fact that there's an easier way

[silk vine]

did i miss something

[iron stream]

Divide numerator and deno with x⁴

[silk vine]

partial fractions seemed obvious

[iron stream]

Not needed

[silk vine]

ohhh

thats very clever

[iron stream]

Do this and sub 1/x -1/x³ = z

[silk vine]

if it works it works though

[mighty meteor]

right

so i remember what i didnt get at this exercise

so what do you guys think the answer is?

[silk vine]

I got c

but i kinda rushed it so not too confident

[mighty meteor]

oh damn you are quite far from the answer

[silk vine]

a pity

ill check my workings

[mighty meteor]

here

so from that it would like the answer is a)

[iron stream]

Can ya do this ?

[mighty meteor]

but then what is very similar to a)?

its e)

and e) is hte answer

[mighty meteor]

man, if you come up with these clever substitutions yourself at 16

then you are a god at maths

[iron stream]

Integrals was my fav

[mighty meteor]

i am not a god at maths tho

[silk vine]

yeah

[iron stream]

But I was 18 then

[silk vine]

i did it and it worked out like crazy

[mighty meteor]

?

[silk vine]

i just have no idea how you spotted it

[iron stream]

I learnt integrals at age 18

[mighty meteor]

didnt u say ure 16? or am i confusing you with someone else?

[iron stream]

I'm 19

[mighty meteor]

oh

damn

never mind then

yea idk, just check what i did

and tell me what i did wrong

[silk vine]

how do you spot these substitutions

well not substitutions but like

[iron stream]

There's no specific answer to address this question.

[silk vine]

that came out of noweher

[mighty meteor]

exactly

like

i dont wanna be mean or anything

[silk vine]

What made you think to divide by x^4

[mighty meteor]

but sometimes it seems like u dont come up with these yourself

welp, i think hes looking them up

but it doesnt matter

i personally could never come up with these subs

because also

its not like they are universal

no

[silk vine]

well in the end that substitution got me the right anser i think

[mighty meteor]

its very specific subs that would only work in THIS specific exercise

[mighty meteor]

can you please check what i did?

this

[silk vine]

idk some people are crazy good at spotting

[mighty meteor]

what answer did you get?

[mighty meteor]

it feels like its inhumane

to spot these things

[silk vine]

e

[mighty meteor]

right

i got a

[silk vine]

the substitution just solves the entire question its crazy

[mighty meteor]

which is e but with the denom reversed

dude, for the love of god

[silk vine]

oh no way i actually got this answer as well

[silk vine]

sure

[iron stream]

You missed a negative sign somewhere

[mighty meteor]

thats for sure

but where

actually no its not about the negative sign

it would have to be (1-x)(1+x) to get 1- x^2

[silk vine]

oh well

thats your negative sign

swap the bottom multiply the entire thing by -1

and that flips the log

[mighty meteor]

huh?

i dont get it

also

this is the solution from the solution book:

which would mean that im right

but then, the solution from the multiple choice book

says its e)

i hate this book

i will burn it

i will do to it what they did in germany in ww2

[silk vine]

it wasnt quiteright

[mighty meteor]

what did i do wrong?

if it was - ln of that

[silk vine]

no i made a mistake when i saidyou could minipulate it

[silk vine]

cause yeah that

[mighty meteor]

then you would get ln (x^2-1)/x^3

which is just wrong

if it were e) there shouldve be a -ln

yea i think im right

the book just made a typo

[silk vine]

I got the sme answer as you the first time i did it

man im gonna check an integral calculator

[mighty meteor]

bro, XOR-11's inhumane formula did us dirty

[silk vine]

integral calculator has same answer as you

mustve been a tyypo

[mighty meteor]

,w integrate [x^2-3]/[x^3-x]

[silk vine]

im sure its right i missed a minus somewhere

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sizzywizzy0_0_32713

[silk vine]

yup

[mighty meteor]

this motherfucker

i will kill wolfram

WHERES THE MISTAKE

I MADE NO MISTAKE

[silk vine]

thats the same answer as you

[mighty meteor]

no

i got x^2 - 1

he got the correct answer

the one from the book

[silk vine]

wild the integral calculator is arguing with wulfrum alpha

[mighty meteor]

is it?

no way

[mighty meteor]

wow

i asked chatgpt too, but his off his goddamn gourd

he god some insane result

[silk vine]

every chatgpt answer is nonsense

[mighty meteor]

hes just a little special

[silk vine]

thats it

differentiating

your answer is right

[iron stream]

It's always advised to consider the 1-x² form

You have considered ln(x²-1) and that's causing the issue

[mighty meteor]

ok so, the correct answer is mine

[iron stream]

Idk the exact reason but my teacher said this !

[silk vine]

yup

[iron stream]

Maybe because then, the function is defined for trigonometric functions

[silk vine]

how did you spot

[iron stream]

What ?

[silk vine]

how did you spot dividing everything by x^4

[iron stream]

Idk bro it just came to my mind 😭😭

[silk vine]

insane

[mighty meteor]

i dont believe you for one second

[silk vine]

like once i see it it makes sense

[mighty meteor]

also, what differentiating site did you find?

[silk vine]

i looked up derivative calculator

[mighty meteor]

wait

both answer

a) and e) differntiate the same

this is some bullshit

[silk vine]

i got it differently when i put in d

[mighty meteor]

it was never d

sorry

i meant e)

[silk vine]

weird

[silk vine]

this is a isnt it

[mighty meteor]

yea

i put both a) and e)

both differentiate the same

but the integral looks different

[silk vine]

wowza

[mighty meteor]

wait

,w differentiate ln(x^3/(1-x^2))

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sizzywizzy0_0_32713

[silk vine]

wulfrum???

oh its fine

[mighty meteor]

,w differentiate ln(x^3/(x^2-1))

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sizzywizzy0_0_32713

[mighty meteor]

alright

i will consider this exercise solved

up for more?

[silk vine]

i dont even know what to make of a and e both being correct

i thought it was uniqu

i guess cause when you integrate its modulus

[silk vine]

yeah

[mighty meteor]

it is

[mighty meteor]

hm? no it isnt

oh wait

yes you are right

[silk vine]

like integral 1/x = ln|x|

[mighty meteor]

so the correct answer is e)

because its modulus

DAYUMMMM

[silk vine]

but also a

[mighty meteor]

no no

it isnt

[silk vine]

why is a not counted

[mighty meteor]

because x belongs to (-1, 0)

[silk vine]

ohhh

amazing

[mighty meteor]

one sec i gotta write it dow

down

here

this is the next one

[silk vine]

okie

oh is this just an integrl

[mighty meteor]

it is

i just didnt get the right answer

these are very easy ones

but ill get back to the super duper hard ones after this one and the next one

[silk vine]

is this one c

[mighty meteor]

well i be damned

it is

ill do it quickly

maybe i get the answer right this time

[silk vine]

cool

[mighty meteor]

the fact that you did it in your head

woww

[silk vine]

didnt do it in my head

[mighty meteor]

alright

[silk vine]

no chance

wrote down very fast

[mighty meteor]

i will have to translate the next one

so hang on

[silk vine]

lovely

its not a

and its not f

[mighty meteor]

i agree

[silk vine]

Im gonna say c and e are wrong

[mighty meteor]

huh?

i actually thought that e might be right

[silk vine]

im sure c is wrong

im not too sure why there would be infinitely many primitives

[mighty meteor]

ill check

[silk vine]

interesting

if a function has one primitive it has infinitely many primitivs

[mighty meteor]

well yea

its because of the constant

c

this wouldnt be the same if it were a defined integral im pretty sure

[silk vine]

maybe b and e then

[mighty meteor]

how about d)?

oh wait

[silk vine]

forgot what that does

im looking it up

[mighty meteor]

it continuous so it does have Darboux's property...maybe

[silk vine]

If f : I → R is a function, and I ⊂ R is an interval, f has
the Darboux property if for any a, b ∈ I, a < b and for any λ ∈ (f(a), f(b)) ∪
(f(b), f(a)) there is c ∈ (a, b) such that f(c) = λ.

aw man

[mighty meteor]

no no

it does have darboux's property

because it says that there is an f'(c) = gama

which is true even if the function is not differntiable on R

it only needs to be continuous and differentiable on an interval

so it does have it

so d) is wrong

[silk vine]

lovely

this darboux property stuff is really coming out of nowehere

[mighty meteor]

yea, idk, its just something the helped with that property that f'(c) = 0

im pretty sure

like he helped Fermat

and then fermat helped Rolle

complicated stuff

doesnt matter

anyway

[silk vine]

darboux is a good chap

[mighty meteor]

we are left with b) or e)

[silk vine]

he made darboux sums

can it not be multiple

[mighty meteor]

sheeeeeeeeeeeeeeesh

[silk vine]

cause b implies e

[mighty meteor]

thank god i dont know em

[silk vine]

its not too bad its just integration

[mighty meteor]

no, it cant ever be multiple

[silk vine]

then im gonna say e

cause b means that at some points the whole functions is negative

[mighty meteor]

oooh, its not b) because it sais on R

[silk vine]

which doesnt make sense

[mighty meteor]

and that would mean that c) is correct too

which isnt

because its not differentiable on R

aha

insane

[silk vine]

why dont you think its e

[mighty meteor]

no, i do think its e

[silk vine]

me as well

[mighty meteor]

i just said that b) is wrong

[silk vine]

oh apologies

is that right?

[mighty meteor]

yea

its e)

im almost finished with this chapter

[silk vine]

hoorays

[mighty meteor]

i can believe it

i almost finished the whole integration part from this book

[silk vine]

in how long

[mighty meteor]

i only got the last 10 turbo hard exercises at the end

[mighty meteor]

idk, like...

4 days

im not sure

it may have been 5

[silk vine]

thats insanely fast

[mighty meteor]

yea, i work 8 hours a day

[silk vine]

lets see these hard exercises

[mighty meteor]

at math

or well, not just math

its math and physics

[mighty meteor]

oh god

i need a break

can i be back in like 30 mins?

ill dm you in private

[silk vine]

sure

Ill probably be off by then

[mighty meteor]

and ill close this chat

[mighty meteor]

oh

you'll be off in half an hour?

[silk vine]

good integrating today though

[mighty meteor]

eh, its fine, we can just do some more tomorrow

[silk vine]

im gonna divide everything by x^4 from now on just in case

bye bye

[mighty meteor]

LMAOOO

same

bye

.close

[pearl pond]

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[pearl pond]

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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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[gusty flume]

Prove the following:
For any weighted, undirected graph $G=(V,E)$:
For any proper subset $U\subset V$ of vertices
and any edge $e={u,x} \in E$ with $u \in U$ and $x \in V\setminus U$
and $w(e)= \min { w({u', x'}) \mid {u', x'} \in E, u' \in U, x' \in V\setminus U }$:
There exists a minimal spanniung tree $T$ with $e\in E(T)$.
And if $w(e)$ is strictly smaller than any other edges between $U$ and $V\setminus U$, then $e$ is included in all possible minimal spanning trees.

[jolly parrot]

Bob Goldham

[gusty flume]

My idea was the following

case distinction:
We consider the graph (V, E\{e})
it is either connected or not connected.
If it is not connected, then e must be in any spanning tree of G whatsoever, including any minimal spanning tree.

If it is connected, then the minimal property of $e$ guarantees that any edge $e'$ between $U$ and $V\setminus U$ has at least weight $w(e)$.
If the weight is larger for all such edges, then the spanning tree of $G'$ is no longer a minimal spanning tree of $G$.

So every minimal spanning tree of $G$ contains at least one edge of exactly weight $w(e)$.

If there is more than one such edge, then we could remove all except (any arbitrary) one of them and still have a minimal spanning tree, since they all connect the same subsets of vertices and have the same weight. Thus the resulting spanning tree must always remain minimal.

[jolly parrot]

Bob Goldham

[gusty flume]

Is this correct?

[pearl pond]

@gusty flume Has your question been resolved?

[pearl pond]

@gusty flume Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[gusty flume]

.reopen

[pearl pond]

✅

[pearl pond]

@gusty flume Has your question been resolved?

[pearl pond]

@gusty flume Has your question been resolved?

[pearl pond]

@gusty flume Has your question been resolved?

[harsh idol]

try pinging helpers

or opening a new help chat

you can ping helpers every 15 mins

[gusty flume]

<@&286206848099549185>

[brave sluice]

you didn't define G'

but i assume G' is (V, E{e})

i don't think the proof is correct

you said this but

isn't this a counter example?

[pearl pond]

@gusty flume Has your question been resolved?

[pearl pond]

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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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• Be polite and have a nice day!

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[exotic crane]

A bucket was sold at 8% loss had it been sold for 56₹ more there would have been a gain of 8% what is the cost price of the bucket?

[cinder thistle]

so

let us the that was p

we have

p(1+8/100)-p(1-8/100) = 56

right?

[exotic crane]

yeah

[cinder thistle]

so

p(16/100)=56

right?

[exotic crane]

uh maybe

im not sure

[exotic crane]

i didnt even did this

should i show my working?

[cinder thistle]

sure

[cinder thistle]

56 = 16/100 x

it should be

[exotic crane]

how?

[cinder thistle]

108/100 x - 92/100 x

(108-92)/100 x

16/100 x

[exotic crane]

uh yeah

oh

i got it

thanks

.close

[pearl pond]

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[pearl pond]

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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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[rapid wagon]

$\lim_{x \to 2^-} \frac{\sqrt{x^2-4}}{|x-2|}$

[jolly parrot]

prograce

[rapid wagon]

Help please

[cinder thistle]

Does not exist, or are you talking in complex numbers?

[rapid wagon]

No, real

Why?

I got 0 idk

[vagrant trout]

x - 2 > 0, x>2, so undefined limit... can't do left side limit only right side

[rapid wagon]

Oh

[vagrant trout]

Because if x-2<0 it becomes 2-x so it's just 2-1.9999999 which approaches 0 and it's undefined

[pearl pond]

@rapid wagon Has your question been resolved?

Channel closed

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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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• Be polite and have a nice day!

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[mighty meteor]

i need help with this

[sharp smelt]

multiply and divide by $\sqrt{e^{x}+1}$

[mighty meteor]

mhm, i think i tried it

let me try again

[sharp smelt]

oops

my bad

[mighty meteor]

thats for sure the right approach

[jolly parrot]

math rocks(wai)

[sharp smelt]

I'm confused

[mighty meteor]

no no, i think that its the right approach, i just didnt do the calculations right probably

let me retry real quick

i think that maybe it would better to multiply by $\sqrt{e^{x}-1}$

[jolly parrot]

Gustavo

[mighty meteor]

because then when you do the sub for $e^{x}$ you get something a little better, maybe

[jolly parrot]

Gustavo

[mighty meteor]

because this is what you get otherwise

[sharp smelt]

huh?

$

$\int \frac{e^{x}+1}{\sqrt{e^{2x}-1}}$ is what you should get

[jolly parrot]

math rocks(wai)

[mighty meteor]

oh yea sorry

i got confused and mixed stuff up

[sharp smelt]

That makes it easier I think?

[mighty meteor]

yea yea, thats what i was thinking too

but then the problem is that you now have u*sqrt(u^2-1) in the denom

[sharp smelt]

You can start with e^x=u

[mighty meteor]

yea thats what i did

thats why theres the u down at the denom

[sharp smelt]

ah, okay

[mighty meteor]

yea

[sharp smelt]

so first split this into the sum of two integrals

[mighty meteor]

so then i thought of breaking the integral into two

[mighty meteor]

alrighty

[sharp smelt]

Think of a way to integrate $\frac{1}{\sqrt{e^{2x}-1}}$ now

[jolly parrot]

math rocks(wai)

[mighty meteor]

idk why you put it back in e^x

i think the u sub looked better

alright

i think i got it

[sharp smelt]

Yeah,but when you let $e^x=u$, you get $e^xdx =du$

[mighty meteor]

i almost did it

[jolly parrot]

math rocks(wai)

[sharp smelt]

not $(e^x+1)dx=du$

righ

[jolly parrot]

math rocks(wai)

[mighty meteor]

and dx = 1/u du

which is great actually

[sharp smelt]

wait, how did you you get $dx= du/u

[mighty meteor]

e^x=u

therefore x=ln u

therefore dx= du/u

[sharp smelt]

ah, that works too

[mighty meteor]

its just a basic trick, idk if you knew that a= e^(lna)

[mighty meteor]

yup

ill show you what the problem is tho

[sharp smelt]

I should hope I know that, I'm doing a UG in maths lol

[mighty meteor]

first of all though

do you understand my handwriting?

[sharp smelt]

yes

[mighty meteor]

because i will show you a picture

[sharp smelt]

looks right to me

[mighty meteor]

mhm

now look at the multiple choice

the problem is

we gotta figure a way out to turn arcsin in arccos

because i checked and d) is indeed the right one

[sharp smelt]

recall the relation between arcsin and arccos

[mighty meteor]

but i have arcsin not arccos

[sharp smelt]

arcsin(x)+arcos(x) = π/2

[mighty meteor]

soooooooo

what does that mean?

[sharp smelt]

well, I think in your working it should be -arcsin(-1/u)

[sharp smelt]

$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$

[jolly parrot]

math rocks(wai)

[sharp smelt]

use this

[mighty meteor]

how?

so arcsin x = pi/2 - arccos x

[sharp smelt]

how is this identity true?

[mighty meteor]

no

how do i use it here?

[sharp smelt]

yes

recall that π/2+C=C here as C is just an arbitrary constant

[mighty meteor]

and also

is arcsin odd or even?

like

is arcsin -x = -arcsin x?

or maybe arccos -x = -arccos x?

cuz otherwise, this is still not doing it

[sharp smelt]

yup

[mighty meteor]

ok

[sharp smelt]

,w is arcsin(-x)=-arcsin(x)

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by math_rocks

[mighty meteor]

ill write it down real quick

1 sec

btw

i got more integrals to solve

you wanna help me more?

or should i make another help channel?

[sharp smelt]

sure

or actually, I have some work to do now

[mighty meteor]

ah oki

no prob

cya

[sharp smelt]

open a new help channel, if I'm done by then,I'll try to help

[mighty meteor]

.close

[pearl pond]

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[pearl pond]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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Channel closed

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✅

[cinder bane]

Refer to #❓how-to-get-help to know ho to get help probably
And also just ask the question, don't ask to ask, this will just consume more time for you.

It would be faster for you and the helpers to just ask the question

.close

[pearl pond]

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[pearl pond]

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[quick sand]

I have a very simple question.
I have got a math idea, which I'm able to prove it, but there is no any application for that, and I don't have any reason for doing this, just curiosity.
I'm an UG student, and the idea is about set theory.
So question is, shall I open this with my professor? or it's very bad idea? Thanks

[open rivet]

mathematicians dont care about applications

if its a problem, and youre able to solve it, why not?

[regal herald]

whats your concern here?

[halcyon venture]

Are you sharing your idea?

[unkempt yacht]

an applied mathematician encounters a problem and uses his knowledge to solve that problem

a pure mathematician creates a solution to a problem that may arise in 10-20 years, not that he gives a shit

[open rivet]

there have been plenty of times where mathematicians discover some maths and then like 100 years later then find it has applications

[vagrant trout]

Well you can share your idea and others might find an application for it.

[iron stream]

@quick sand Read A mathematician's apology by gh hardy

[pearl pond]

@quick sand Has your question been resolved?

[quick sand]

Hi, thanks for answering. Well there is no problem too. Just playing with math stuff, since it's fun

[quick sand]

Nothing really(yet)

I haven't constructed my idea into papers properly, but I do it. And I'm able then to prove and setup relations and properties of my stuff.
So if it's a normal thingy, then I'm going to open it with my prof soon. I'm just too scared my prof finds it very nonsense since there is no application or reasoning for that, although he's a pure math guru as well.

[pearl pond]

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[gusty flume]

$x^3-3x^2+3x-7=3p$

[jolly parrot]

Bob Goldham

[gusty flume]

I want to solve for x, p is a variable

Is there any simpler way than the Cardano formula?

$x^3-3x^2+3x-(3p+7)=0$

[jolly parrot]

Bob Goldham

[worthy lance]

Do you know (a-b)^3?

[gusty flume]

=(a²-2ab+b²)(a-b)
=(a³-2a²b+ab²)-(a²b-2ab²+b³)
=a³-3a²b+3ab²+b³

=a³-b³-3ab²+3a²b

that?

[worthy lance]

What do you have from that in your original

Also

Is -b^3

3a^2b and -3ab^2 too

In any case

You have almost all

For (x-1)^3

Just split -7 into -1-6

[gusty flume]

(x-1)³=
x^3-3x^2+3x-1

[gusty flume]

so this term

[worthy lance]

So (x-1)^3-6

[gusty flume]

is just (x-1)³-6=3p

[worthy lance]

= 3p

[gusty flume]

is there a name for this formula? like the binomial formula for (a-b)²?

[worthy lance]

(x-1)^3=6+3p

Cubic root both sides

Cube of a binomial

[gusty flume]

So i get

$x=\sqrt[3]{3p+6}+1$

[jolly parrot]

Bob Goldham

[gusty flume]

alright, thanks!

.close

[pearl pond]

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[pearl pond]

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[stiff mauve]

can someone help

[stiff mauve]

ill try to explain where i'm at right:

i know column [1 1 1 1 1 1] transpose is a solution

i now need to find solutions where ax = 0

to get the set of all solutions of the system Ax = [ 1 1 1 1] transpose

[stiff mauve]

i make augmented matrix with [ A | 0 ]

i dont know how to continue

anyone?

[tropic saddle]

after you reduce [A | 0], what do you get

[stiff mauve]

i begin by saying x1, x4 x5 are free variables

[tropic saddle]

and then?

[stiff mauve]

x2 = x4 - x5

x3 = -x4

x6 = 0

x1 = x4 = x5 = free

ohhh righttt

your presence was enough thanks i understand it now

[warm yew]

heys

(1,1,1,1,1,1) + b(0,1,-1,1,0,0) + c(0,-1,0,0,1,0)

a,b,c are real numbers

whatever

[warm yew]

should be correct idk

[warm yew]

Where tf the 1 come from

[stiff mauve]

we need to find every possible solution of ax = [1, 1, 1, 1] right

one possible solution is when x = [1 1 1 1 1 1]

[warm yew]

oh yeah

[pearl pond]

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[plush sundial]

im getting lambda as 23

but in my key its given 28

[versed mica]

!show

[pearl pond]

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

[plush sundial]

here is my attempt

ig u wont understand what i wrote

[versed mica]

it is hard to follow

[plush sundial]

ok ill explain

the given eqn of the normal is y=2x-lambda

slope of normal is 2

so slope of tangent is -1/2

and i differentiated the parabola eqn and got dy/dx as slope

assumed the point of intersection as y itself (got too lazy lol)

and equated the slopes

from there i got y as (-6)

feel me?

[versed mica]

yep

[plush sundial]

aight after that i put the value of y in the parabola eqn

got x as 11

so the point is (11,-6)

to where the tangent is drawn

[versed mica]

yea 11,-6 is good

[plush sundial]

then i used y-y1 = m(x-x1)

for eqn of normal

x1,y1 = 11,-6

[versed mica]

what

just use the equation of the normal
line

y = 2x - lambda

y = -6

x = 11

-6 = 22 - lambda

lambda = 28

[plush sundial]

holy

but why was my method wrong?

like again equating it

[versed mica]

what is this for

[plush sundial]

like i got the POI and i have the slope m

x1,y1 = 11,-6

oh shit

ye i got it

man brain crashed

thx tho for confimation

.close

[pearl pond]

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[plush sundial]

.reopen

[pearl pond]

✅

[plush sundial]

i hv another doubt

im a lil confused in this one

@versed mica srry for ping

[vagrant trout]

first complete the square, get the radius

then calculate the distance between the center of the circle and the point you got (1,3)

call it d for distance

[median monolith]

hello guys sorry i need help about a math thing

[vagrant trout]

then also calculate the distance between the point (1,3) and any point on the circle

[plush sundial]

occupied for a while brotha

[vagrant trout]

call it D

[plush sundial]

how do i do that?

[vagrant trout]

then max distance is

d+D

[median monolith]

what the hell bro

[plush sundial]

wont it be d + radius?

[median monolith]

just asked everyone not just you :>

[vagrant trout]

yeah

d + radius

as any point on the circle

is just the ending point of the radius

[plush sundial]

oh ye got it

d=7, radius = 4
ans 11

ggs

[vagrant trout]

yes

[median monolith]

rookies

btw i am working on uhh some kinda integrals that has circle in middle of em

[vagrant trout]

wait

open a new one

in available

go ina ny help channel

[vagrant trout]

in available

[plush sundial]

ill close this

.close

[pearl pond]

Channel closed

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[warm current]

You didn't have to.. But it did get a little cluttered in here...

[pearl pond]

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[jolly parrot]

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

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[rancid steppe]

√x^2 -6x +25 + √y^2 +8y+ 41 <= 9 \newline √x^2 - 6x + 9 - 9 +25 + √y^2 + 8y + 16 - 16 + 41 \newline √(x-3)^2 + (y +4)^2 <=9$ what now

come on are you kidding me right now? I waited 30 mins for a channel

[dusk horizon]

Missing a dollar sign?

[rancid steppe]

Yes but the compiler did not made it as it is supposed to be

do i open a new help chat now?

[dusk horizon]

Just try your message again here

[leaden wadi]

Perhaps don't delete the original message. 🤷‍♂️

[rancid steppe]

I am not trolling?!!

[leaden wadi]

They deleted the original message which is why the channel was closed.

[spare lark]

30 min ?!

[rancid steppe]

${√x^2 -6x +25} + {√y^2 +8y+ 41} <= 9 \newline {√x^2 - 6x + 9 - 9 +25} + {√y^2 + 8y + 16 - 16 + 41}$

[jolly parrot]

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

[rancid steppe]

why cant i type √

[spare lark]

Its better that you open a new one they're free now

[leaden wadi]

Use \sqrt{}

[rancid steppe]

i am opening a new one

#help-27

[pearl pond]

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[brave sluice]

are groups of prime order cyclic?

[brave sluice]

or are groups of prime order abelian?

i think they are but i'm looking for a theorem in my book

or am i misremembering something?

[novel apex]

I know that groups of order p^2 are abelian

maybe it is the same for groups of order p

[fluid axle]

both

[brave sluice]

thanks

[main oxide]

if you can prove they're cyclic, then you're good cause cyclic groups are already abelian

[fluid axle]

cyclic implies abelian

[brave sluice]

what is the argument?

for prime -> cyclic/abelian

my book has a chapter on cyclic groups

[main oxide]

you might need to use lagrange's theorem

[brave sluice]

the order of an element divides the order of the group

therefore an element other than the identity generates a subgroup the same size as the group

[novel apex]

so every elements can be written as power of other elements

then you can conclude

[brave sluice]

oh i found it as a corollary to the theorem of lagrange

every group of prime order is cyclic

thanks folks 🙏

.solved

[pearl pond]

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[regal heath]

are there any videos about homogeneous and non homogeneous recurrence relations with explanations of why the "formulas" work?

[pearl pond]

@regal heath Has your question been resolved?

[pearl pond]

@regal heath Has your question been resolved?

[pearl pond]

@regal heath Has your question been resolved?

[pearl pond]

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[charred lynx]

idk how to do this

[limpid lily]

I'm not sure exactly what method they want you to use, but one method is to use the fact that the derivative (here, g') is the slope of the original function at that point.

So, when x is 3, g'(x) gives you 3, so the slope is 3 there.

[charred lynx]

eah

its

approximation

tangent line approximation

[limpid lily]

Right, so the formula from algebra for the slope is (y0 - y1)/(x0 - x1) or something like that and the slope is 3.

[charred lynx]

Ooooooh yeahhh wait

I see

i get it now.

thx.

.close

[pearl pond]

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[limpid lily]

You're welcome.

[pearl pond]

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[sharp smelt]

Determine if it's possible to have a cauchy sequence with an unbounded sub-seqeunce.
\
I was thinking $a_n = \sum_{i=1}^{n} \frac{1}{i}$

[jolly parrot]

math rocks(wai)

[west sapphire]

is that a cauchy sequence?

[sharp smelt]

Isn't it?

I see

[west sapphire]

do you know the theorem that (in R or C)
cauchy <==> convergent

[sharp smelt]

yeah it isn't

oh

I forgot the backward implication

oops

Thanks

.close

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[mellow idol]

hi, i was solving a question on functional equations and to prove the functions injectivity i assumed f(af(0))=f(bf(0)) and then got this implies a^2=b^2. but then this implication would mean f(0)=0 which would make my initial assumption f(0)=f(0) does this make it invalid?

[warm current]

!xy

[pearl pond]

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.

[mellow idol]

<@&286206848099549185>

[stable dune]

i mean the initial assumption that f(0) = f(0) is valid

you basically just showed that your assumption of f being injective doesnt fail in this regard

[pearl pond]

@mellow idol Has your question been resolved?

[mellow idol]

huhh

i mean f(0) = f(0) need not be assumed right

i assumed f(af(0))=f(bf(0)) and got that the functions injective but when i use injectivity i also get f(0)=0

im asking does f(0)=0 invalidate assuming f(af(0))=f(bf(0)) to prove injectivity

[mellow idol]

can someone tell me if my proof is valid or not? cause if it is it trivialises the question

[main oxide]

I don't really see or understand your proof

[pearl pond]

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[harsh nymph]

Hello can someone give me some resources for prepare for math competition? or some book for proof? I'm currently in high school year 12

[sharp smelt]

For proof , book of proof ( yes, that's the name) is a nice book

[harsh nymph]

is it beginner friendly?

like not too advanced concept

[sharp smelt]

For the most part, yes

[harsh nymph]

oh ok i will check it out

[stable dune]

oh i see

[pearl pond]

@harsh nymph Has your question been resolved?

[cinder thistle]

what is the syllabus?

[pearl pond]

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[quick star]

stupid question but percent increase can be both additive and multiplicative?

[pure rapids]

wdym

[quick star]

increase by 20% and then a further 20% can be additive or multiplicative.

additive: 100% + 20% + 20% = 140%
multiplicative: 1 * 1.2 * 1.2 * 100 = 144%

[errant fable]

this is a statement not a question

oh wait

[quick star]

what

where am i making a statement?

[nocturne pier]

https://tenor.com/view/bunnies-what-confused-meh-gif-12007368131756137942

like can you give us the exercise or something?

[pure rapids]

Can be both add and times i think

[bitter herald]

its like

the difference between compounding and simple interest

to give you an analogy

the multiplicative one "adds on" to the previously applied percentages

whilst the additive one is always a linear increase based on the principal amount

So yes, percentage increases can be additive or multiplicative, depending on the context and the assumptions made about the base value

[pearl pond]

@quick star Has your question been resolved?

[pearl pond]

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[prisma elk]

quick question: if z = (1/ i*x )
lim x->0+ arg(z) = ?

[pearl pond]

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[ebon burrow]

<@&286206848099549185>

[regal plume]

@pearl pond

[ebon burrow]

Hello

Pls help

<@&286206848099549185>

[sacred crest]

Don’t spam the helpers ping

[ebon burrow]

Ok sorry

😦

@sacred crest could u help me ig?

[pearl pond]

@ebon burrow Has your question been resolved?

[cinder thistle]

so where did you get till yet?

[ebon burrow]

Couldn't even start

@cinder thistle

[cinder thistle]

try constructing a square

[ebon burrow]

Ok wait

Doin

[sacred crest]

How come you couldn’t start?

[ebon burrow]

I did not understand the first part

Like how to proof.

Give me some hint please

[sacred crest]

Just try things out

[ebon burrow]

Not working for 3x3

[sacred crest]

How about 5x5 then

[ebon burrow]

Do you know that's 5! Many combinations

About 120

I'm not to try ot

Is there, say any suitable algo for it?

[sacred crest]

Idk, could be. It looks like you’re supposed to figure out such an algorithm on your own though, cause you’ll learn a lot more that way. Unless this is supposed to be easy homework

[pearl pond]

@ebon burrow Has your question been resolved?

[pearl pond]

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[sharp smelt]

If $(x_n)$ and $(y_n)$ are cauchy seqeunces then prove that $x_n+y_n$ is a cauchy sequence, without using the cauchy criterion or the algebric limit theorm

[jolly parrot]

math rocks(wai)

[sharp smelt]

I'm confused

so here I cannot use the defn of a cauchy seqeunce in my proof?

The proof I had in mind was $\abs{x_n - x} + \abs {y_n -y} = \abs{x_n-x_m + x_m -x} + \abs{y_n - y_m+y_m-y}< 2 \varepsilon \implies \abs{x_n-x_m} + \abs{x_m-x} + \abs{y_n-y_m}+ \abs{y_m-y} < 2\varepsilon$.

[jolly parrot]

math rocks(wai)

[sharp smelt]

Now as $\abs{x_m-x}$ and $\abs{y_m-y}$ convergre, i replace, both with $\varepsilon/2$, to obtain $\abs{x_n-x_m} + \abs{y_n-y_m} < \varepsilon$

[jolly parrot]

math rocks(wai)

[sharp smelt]

That would tell us the series is a cauchy seqeunce

is this fine

[pearl pond]

@sharp smelt Has your question been resolved?

[sharp smelt]

<@&286206848099549185>

.close

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[sleek junco]

can someone explain how n^-n come

[sleek junco]

at the third equal sign

[open rivet]

they are doing $(n+1)^{-n}=\left(n\left(1+\frac1n\right)\right)^{-n}=n^{-n}\left(1+\frac1n\right)^{-n}$

[sleek junco]

oh

disgusting but oh welp

thx for the help

[jolly parrot]

Bonk

[sleek junco]

.close

[pearl pond]

Channel closed

Closed by @sleek junco

Use .reopen if this was a mistake.

[pearl pond]

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[quick star]

,, a \frac 1b

[jolly parrot]

mmmm7

[quick star]

is this mixed fraction or product?

i know something like $2 \frac 12$ is mixed fraction but like is it the same with variables?

[jolly parrot]

mmmm7

[safe prairie]

thats just a/b

thats just a normal fraction

[quick star]

why is it not mixed fraction?

[safe prairie]

but if you insist it be written that way yeah its mixed

for mixed fraction you see them as 2 + 1/2

for variables youd just multiply any adjacent thing

[quick star]

yes mixed fractions means that

but that's not notation for mixed fraction, is it

[jolly parrot]

mmmm7

[safe prairie]

ofc 2 + 1/2 is a wrong notation

[safe prairie]

but unless you tell your readers this is a mixed fraction youd expect everyone to treat it as a/b

[quick star]

depends on context?

my teacher wrote that and expected us to know it's a mixed fraction

[safe prairie]

a + 1/b?

[safe prairie]

this or (ab+1)/b would be much more common than mixed fraction notation

[quick star]

yeah well they wanted me to interpret it that way

anyway thanks

i won't make a big deal out of something trivial

😭

.close

[pearl pond]

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[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rancid steppe]

$ax - bx + ax^2 - bx^2$

[jolly parrot]

Simon James B

[rancid steppe]

I did $x(a-b) + x^2(a-b) \newline (a-b)(x + x^2)$

[jolly parrot]

Simon James B

[rancid steppe]

but it seems that this was not right

[open rivet]

it should be correct

[warm copper]

Factorise x again

x(x+1)(a-b)

[open rivet]

ah, yes

[rancid steppe]

ah i see it

$x(1+x)(a-b)$

[jolly parrot]

Simon James B