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[vocal sleet]

@upbeat maple Has your question been resolved?

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[patent nymph]

consider: $$ \frac1{a(a+1)} = \frac1a - \frac1{(a+1)} $$

[twin meteor]

rysrobrgldvoelr👻ep>vneae=u

[patent nymph]

you don’t need that for this one

[patent nymph]

only this, and then expand it out and cancel infinitely many things

Yes

Not all

There’s 1

No

so, can you find the answer?

it means

something like this:

$$ \frac11 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \cdots $$

[twin meteor]

rysrobrgldvoelr👻ep>vneae=u

[patent nymph]

where you can rearrange the terms, and then get out a different result

Here, this is from Wikipedia

A series is conditionally convergent when you can rearrange its terms to get a different sum

The picture I sent shows how this can be done

[silk garden]

I don't know if I'm allowed to input, and apologies if I missed this somewhere, but wouldn't it be easier to structure it as a power series and evaluate the solution using series tests to determine what it meets?

Oh nvm I see you were explaining telescoping

[patent nymph]

There is no reason to structure it as a power series, it’s already telescoped

[patent nymph]

You’re always allowed to input

[silk garden]

So based on the telescoping wouldn't it have a defined sum and be convergent?

And wouldn't conditionally convergent mean that in one kind of test it's confirmed to converge but in another it diverges

[patent nymph]

They were asking about what conditionally convergent means

[silk garden]

Oooh okay

[patent nymph]

I took that to be a question they had after they solved the problem

[silk garden]

What did you read it as

[patent nymph]

you may have seen a definition like “conditional convergence means that if you take the absolute value of all the terms and add them up, then the result diverges”

turns out, these two are equivalent

a series which is convergent, but whose absolute value series is divergent, can always be rearranged to give any value you want

and vice versa

No, (1/2)log 2 is different from this problem that you sent

It was just an example

It’s a multiple choice question, I don’t think you’ll need to show it

This series is absolutely convergent for x = 1

not conditionally convergent

so d is not the correct answer

because its absolute value series doesn’t diverge

If you like, but there is no need to, because all terms are positive

[silk garden]

Since this question was just going off telescoping, and telescoping is always convergent to a sum you shouldn't need to worry about it being conditionally convergent

If say you were working with something that relied on the alternating series test, then you would have to check for that conditional or absolute convergence

[patent nymph]

at first, there was no negative sign

[silk garden]

[vocal sleet]

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[vocal sleet]

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[wide sundial]

Hello guys, not sure if i'm being dumb but i wasn't even to justify one quantity is bigger than the other

[twin meteor]

kanna

kanna

[brisk moss]

[wide sundial]

[brisk moss]

have you tried playing with some specific numbers?

[wide sundial]

yeah i think r is bigger by testing some numbers, but is there an algebraic way of showing this?

also n = 0 they're equal right i'm confused lol

[brisk moss]

ok here is something to think about

for large n, no matter what U and G are, (U+n)/(U+G+2n) is going to about n/(2n) = 1/2

[wide sundial]

oh yeah just limits stuff

[brisk moss]

and U/(U+G) can be less than 1/2 or greater than 1/2 depending on U and G

so that's enough for me

[median crane]

Real kanna helping fake kanna

[wide sundial]

i'm fake because i suck at mathsmh

[wide sundial]

but the relationship can't be determined then right?

[brisk moss]

yep

[wide sundial]

i think the answer is r is always bigger

iirc

or p is always bigger

it's one of those

[brisk moss]

[wide sundial]

okay wait i think i see

it says "club had more undergraduate students than undergraduate students" so i guess P < 1/2, right?

and n asymptomatically approaches 1/2 in the limit

for large n

so b?

[brisk moss]

ah

[wide sundial]

okie thank you kanna!

apparently i'm not kanna

[brisk moss]

yes i overlooked that

[wide sundial]

btw idk if it's english/math anymore but 😭 can n be 0?

it does say "n more" but they've been messing with me before and taking conditions where n = 0 and stuff

[terse forum]

Two kannas?

[wide sundial]

@brisk moss is the real kanna

[boreal remnant]

Two kannas?

[vocal sleet]

@wide sundial Has your question been resolved?

[brisk moss]

so it looks this isn't necessary: what happens is U/(U+G) is less than 1/2 and (U+1)/(U+G+2) > U/(U+G)

and picking larger n's can be viewed as repeated (U+1)/(U+G+2) > U/(U+G)

[wide sundial]

wait why did you pick different n's

aren't they supposed to be the same n's

unless the rationale is different than what i'm understanding

[brisk moss]

well what i'm saying is (U+1)/(U+G+2) > U/(U+G) is true for any positive integers U,G with G > U

and that also gives us (U+2)/(U+G+4) > (U+1)/(U+G+2)

and so on

[wide sundial]

owhh that's interesting

[brisk moss]

so (U+n)/(U+G+2n) > U/(U+G) for any U,G,n

[wide sundial]

wait so this is equivalent to showing that:

x/y < (x+t)/(y + v)

for positive x,y,t, and v(?)

[brisk moss]

we shouldn't need 4 variables

just (U+1)/(U+G+2) > U/(U+G)

and we can write G = U + x with x a positive integer or smth

to maybe encode the G > U condition nicely

but still 2 variables

[wide sundial]

okie okie i get it

thanks~

btw what about the asymptotic stuff

[brisk moss]

unnecessary

[wide sundial]

for that we can't show for smaller values of n though, right?

[wide sundial]

oh

okie okie tyty

[vast shale]

[brisk moss]

fake snow walking around sullying us

[wide sundial]

.close

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[cedar ingot]

pls help me with this

[velvet torrent]

AD and AC are the diameters.
they split their respective circles into semi-circles. correct?

not, you know the property. the angle projected by the ends of diameter, on any other point on the semicircle/circle is 90deg

so, angle ABD is 90

and so is ABC

now, if ABD is 90, and ABC is also 90, they sum up to be 180.
this is the case for a straight line

thus DBC is a straight line.
and B lies on the line segment DC

[vocal sleet]

@cedar ingot Has your question been resolved?

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[cedar ingot]

thx

[vocal sleet]

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[sudden compass]

been at this for 30 minutes, all i could reach was

$- \frac{63w +89}{380}$

[twin meteor]

ItzKraken

[minor plaza]

What's the answer

[sudden compass]

idk

[elfin surge]

these q are pain in a**

[minor plaza]

Agreed

[sudden compass]

i know that

[patent nymph]

what is this equal to?

[elfin surge]

a lot of times wht i end up proving

[sudden compass]

x+y+z+w

[elfin surge]

is 0=0

[patent nymph]

Looks like you can use linear algebra to solve this

[sudden compass]

btw after this i do wanna know if theres a general algorithm for this

[minor plaza]

I am trying to write every variable in terms of w

[sudden compass]

gaussian elimination?

[patent nymph]

Put all the variables one side

[sudden compass]

btw me is guessing this is also thru LA

[patent nymph]

There is one

[sudden compass]

and do what?

[patent nymph]

I think it is

[patent nymph]

then eliminate x from all equations but one

I’ll demonstrate 15 min later

[sudden compass]

oh okay 👍 lemme try till then

[sudden compass]

i got 3 equations in total so u want me to eliminate from 2 correct?

[patent nymph]

should be

[sudden compass]

aight

[patent nymph]

These are my workings

all I did was eliminate w from 2 equations, and then eliminate z in one of those

then get everything in terms of x

so x + y + z + w is now all in terms of x

[sudden compass]

well i got it in terms of w (another free variable) the problem is the answer has to be an exact number

[patent nymph]

consider the coefficients

they all cancel out

[regal cradle]

get equations for any three variables, and then use cramers method

[patent nymph]

no need to use cramer

[patent nymph]

my answer was 5

[sudden compass]

how tho..?

[sudden compass]

what

[patent nymph]

,rotate

[twin meteor]

Couldn't find an attached image in the last 10 messages.

Dec 27, 2023, 11:22 AM

[patent nymph]

,rcw

[merry python]

sup kraken

[patent nymph]

I completed my working at the bottom

[sudden compass]

oh in ur work the x cancel out

ah in mines they didnt

must've messed up somewhere then

[sudden compass]

eyy b-eard hows the holidays going

me is fine

[patent nymph]

I think this algorithm is called gaussian elimination

[sudden compass]

yeah thats what u did

U write the coefficients as a matrix

then draw a bar

then the RHS's respectively

and then u eliminate the variables 1 by 1

[patent nymph]

yep

[merry python]

wait

this is easy af

with a bit of intuition and guess

[vast shale]

😭

[merry python]

let me show it

$I\ \to\ 6x+3y-z-w=9\ $\$II\ \to\ 3x-8y-10z-8w=16$\$III\ \to\ 4x-3y-6z-5w=10$\$I+II-2\left(III\right)$\$6x+3y-z-w+3x-8y-10z-8w-2\left(4x-3y-6z-5w\right)=9+16-2\left(10\right)$\$x+y+z+w=5$

[twin meteor]

B-eard

[merry python]

I just noticed the coefficients and tried to form a relation

[sudden compass]

...

well i dont have much intuition sadly

in these questions

[merry python]

Well elimination stuff that rysrobrgldvoelr talked about is quite lengthy

and going for a sort of calculated guess is a better bet

[sudden compass]

how does a person even begin to notice that (1 + 2 - 2(3))

[merry python]

not exactly hit and trial

but kind of yeah

GUESS

[sudden compass]

hmm well I kinda feel ashamed for not noticing this

[patent nymph]

the advantage is that you can start doing it as soon as you see the question

[merry python]

yeah I must agree on that

[sudden compass]

I have done many of these in 3 vars

this was my first in 4 vars

[merry python]

But I'm not a fan of writing much so I'd kind of always prefer tricks (if any)

[patent nymph]

sometimes there’s no quick trick

[merry python]

yeah

[sudden compass]

its like uhh a failsafe if u cant notice the correct thing quickly or are just too bad at it then use it

right?

[sudden compass]

hmm

well tyvm

[merry python]

It is some sort of an olympiad question?

[patent nymph]

finally I’m able to solve the problem in a green name channel

[merry python]

uh..frosst

[sudden compass]

yeah we had a 10 question exam in school with olympiad questions this was the only one i skipped

[sudden compass]

eh? hes solved somethin for frosst before?

[merry python]

NO

[sudden compass]

what NO

[merry python]

Like the kind of problems he posts

[sudden compass]

oh yeah those are always a headache

[sudden compass]

i tend to post extremely easy questions imo

[merry python]

I'd feel so great about myself if I am even able to read it completely

[sudden compass]

lollllll

i tried one of them before it was some algebra+calc question with summations I ended up wasting 2 hours 😂

in the end, snow-sama helped frosst

the solution was obviously beyond my knowledge

well tyvm both of you

.close

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[wide sundial]

hi guys, dumb question but i was trying to find the range of x + 1/x

[wide sundial]

and i saw this person doing it in a unique way without inverses

i don’t understand the f(x) = … part

[vast shale]

do you know what the rearrange inequality is?

[wide sundial]

oh lol is it a thing

i thought he meant by rearranging

[vast shale]

yes it is a thing

https://en.m.wikipedia.org/wiki/Rearrangement_inequality

[wide sundial]

i still don’t get it in first glance

okay i think i see lol

this is a weird thing

never knew math worked like that where u can have inequality relations by shuffling stuffs

.close

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[visual frost]

If I had to prove that a set is a vector space(over a field,ofc),can i avoid proving the 8 axioms by just proving it’s a vector subspace?

[visual frost]

Please pin me ,if u respond

[hard atlas]

if it is a subset of something that you already know is a vector space, then yes

@visual frost

[visual frost]

So,when can’t I apply this?

What’s an instance of having to prove a set is a vector space by those 10 axioms?

[hard atlas]

well for example F^n or the space of all functions

[visual frost]

So why couldn’t I prove it’s a vector space by proving it’s a subspace of itself?

[hard atlas]

because then you never prove all the other axioms like associativity etc

the only reason you can leave them out when proving subspaces is that you already have proved before that they hold in the big space

[visual frost]

But if i had F^2 ,this wouldn’t be the case,correct?

[vast shale]

it needs to satisfy all axioms of a vector space

[hard atlas]

you would still need to show it for F^2

[vast shale]

you must prove all the vector space axioms. If you're proving that a set is a subspace, you still need to show that it satisfies these axioms, as the subspace itself must be a vector space

[visual frost]

So when don’t I have to use all those axioms,when i have a given vector space,of which another subset would be its consequent subspace?

[visual frost]

Well,yeah,but by proving it’s a vector subspace i would implicitly arrive to the conclusion that it’s a vector space

[hard atlas]

you can only prove that something is a subspace if it is a subset of something you already know to be a vector space

[hybrid flicker]

How to show that [insert any possible set] is a vector space fallacy :

• Step 1 : show that it's a subset of itself
• Step 2 : it is indeed stable by any operations you want
• Step 3 : profit

[vocal sleet]

@visual frost Has your question been resolved?

[visual frost]

So,this should be given in the exercise,right?

In order to use this proof

[visual frost]

Yeah,it doesn’t sound right to me either

[vocal sleet]

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[tepid basin]

and then one tails. Give the beta distribution that describes this. Use
integration to determine the probability that the true rate of flipping
heads is between 0.4 and 0.6, reflecting that the coin is reasonably fair.```

[vocal sleet]

@tepid basin Has your question been resolved?

[flat whale]

!show

[vocal sleet]

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

[tepid basin]

Okay, please wait!

[flat whale]

?

[tepid basin]

I think this is the solution! But I don't know how right I am.

[tepid basin]

.close

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[still dune]

So the answer key says unbounded for the equation

but desmos says otherwise

the limit actually seems to be (x -4)?

can someone explain which one is correct?

[gaunt sparrow]

x-4 is unbounded at infinity,

The limit at infinity does not directly tell you about affine asymptotes.

It just asks what the function approaches, and if the function just increases then it's unbounded.

[still dune]

so, although a slant asymptote might sound like it's bounded, it's actually unbounded?

[gaunt sparrow]

Yeah. All a slanted asymptote tells you is that your function looks like a line for large values of x.

But in any case I think it'll be unbounded.

Because if it approaches a line, that line has to go to -infinity or +infinity

[still dune]

ah actually that makes sense

i wasn't too familar with the definition 'unbounded' but now i seem to understand better

thank you

[gaunt sparrow]

No problem!

[still dune]

.close

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[scenic ravine]

Not sure what I have to do here, it's obvious 1/a+1/b=1/2 , but how do I find the line equation from that

[scenic ravine]

Just a hint please

I also got ab=2a+2b

but I don't suppose that helps too much

nvm , just realised a=2b/(b-2)

[vast shale]

uhm

[scenic ravine]

that solves it I think

[scenic ravine]

where a,b are the x and y interecepts

[mystic remnant]

hi @scenic ravine do you want a hint

[scenic ravine]

yeah, am I on thr right track?

[mystic remnant]

hmmm

[scenic ravine]

finding a relationship between a and b should enable me to find the line equation , right?

[mystic remnant]

why not use y=mx plus b

[scenic ravine]

because they have said the reciprocal of the mean of the intercepts is 1/4

so I'm using x/a + y/b=1

[mystic remnant]

calculate the slope with y2-y1 into x2-x1

then you have the y intercept

then you can subsittute

m = (2 - 1) / (2 - 1) = 1/1 = 1.

1 = 1(1) + b
1 = 1 + b
b = 0.

[scenic ravine]

where did you get the 2 from?

[mystic remnant]

the answer is B

its 1,1 2,2

For stone C (4,4):
4 = 4(4) -> 4 = 4. So stone C is on the path of the man.

[scenic ravine]

the answer key says D though, guess it;s wrong then

thanks

[mystic remnant]

no i am wrong i read the question wrong so sorry mb

[scenic ravine]

<@&286206848099549185> , could I have a hint please

I'm stuck with $\frac{1}{a} +\frac{1}{b}= 1/2$

[twin meteor]

Why am. I here

[mystic remnant]

add it

[scenic ravine]

add what?

[mystic remnant]

this is your question right ?

[scenic ravine]

yeah 2ab=a+b is what I get, how does that help?

[mystic remnant]

no wait you have done something wrong

[scenic ravine]

I also know $\frac{x}{a} +\frac{y}{b}=1$ but I don't suppose that helps too much

[twin meteor]

Why am. I here

[scenic ravine]

what?

[graceful moon]

Here is how to add it:

• Cross-multiply 1 and b
• Cross-multiply 1 and a
• Add the two products together, which we get a+b as our numerator
• Multiply the two previoud denominators, which we get ab

Build our fraction with the parts we have, and we get a+b/ab

Oops typo

[mystic remnant]

it should be upon 2

your eq is wrong

[graceful moon]

And like the problem said, The arithmetic mean of the reciprocals of the intercepts of the line on the coordinate axes is 1/4

That's the start that i forgor to add hihi

[mystic remnant]

its 1 upon a plus 1 upon b upon 2 equals to 1 upon 4

that becames 1 /a plus 1/b which is 1/2

and you get one more equation

can i sedn you a phot @scenic ravine of my solving

[scenic ravine]

just send it in this chat, If you don't mind

[mystic remnant]

ofcv

when you solve 1 and 2

you get coordinates of B

[scenic ravine]

huh, but aren't h and k unknowns too?

[mystic remnant]

wait there isanother way this is too direct

take the line as y equals to mx plus c

so the x intercept now is -c/m

and y intercept is c

add the intercepts while dividing by 2 equals is 1 upon 4

-m/c plus 1/c the whole divided by 2 is 1/4

2(1-m)=c

so now the y=mx plus 2 (1-m)=c you will get (y-2)-m(x-2) so it is 2,2

[scenic ravine]

Just looked at the solution . How's this obvious?

[mystic remnant]

look at my sol above its diffrent yet ssame

[scenic ravine]

hmm, OK. Thanks

[mystic remnant]

do you understand ??

[scenic ravine]

Got it, thanks!

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@scenic ravine Has your question been resolved?

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[quaint breach]

Need help with this please

[quaint breach]

I've done this so far but I'm stuck now

[vocal sleet]

@quaint breach Has your question been resolved?

[vocal sleet]

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[hasty tiger]

so i am studying in class 10 icse so in geometry which i find very tough so how to know how to do construction because then we can randomly do anything and do prove anything but if we do that it becomes wrong

[hasty tiger]

so i am studying in class 10 icse so in geometry which i find very tough so how to know how to do construction because then we can randomly do anything and do prove anything but if we do that it becomes wrong

...

sed life

[vast shale]

don t spam

[hasty tiger]

ok then

[lavish sluice]

There's a book called geometry volume 10

It contains various insights

[hasty tiger]

ok

[lavish sluice]

About why construction is being done

[hasty tiger]

ok

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[fast geyser]

Is this accurate?

[twin meteor]

Samuel

[loud walrus]

Check if it's correctly translated, I'm not sure I understood your writting.

[fast geyser]

I dont think it is, sorry for my poor writing

.close

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[pale wagon]

i tried to first iintegrate on z

which results to (5x+6y)/7

if im not wrong

but then im not sure how to continue

it looks like i should use polar coordinates

[mystic remnant]

hi

[pale wagon]

hi

but none of the answer has pi on it

which makes me suspicious

[vocal sleet]

@pale wagon Has your question been resolved?

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@pale wagon Has your question been resolved?

[river wagon]

I also fell it's a bit wierd there is no pi.

I believe the x-part integrates to 0, but the y part should be positive I belive.

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[vestal fractal]

so im writing quadratic equations for this project and i would like for them to be cooler, and be kinda bumpy

[vast shale]

Why means?

[visual oracle]

"be kinda bumpy"

could you explain what you mean by that

[vestal fractal]

Ok so like hmm

say between x = 20 and x=40 i would like for there to be a temporary increase in value before going back down to the rest of the line

or what if i wanted to write more complex lines then this

how do i do it

[visual oracle]

to the rest of the line you mean the rest oft he parabola?

[vestal fractal]

yes

i am stuck in parabola

i dont want perfect parabola

[silk osprey]

so u don’t want it to be a quadratic?

[vast shale]

Could you draw in MS paint what you want exactly?

[vestal fractal]

this is more what i was looking for

[silk osprey]

parabolas can only have one extrema

[vestal fractal]

so

[silk osprey]

if that’s what u mean by bumpy

it can only have one "bump"

[vestal fractal]

my maths education stopped at algebra 1

quadratic equations last thing i learned in school

i do calculus but only in computer words

[visual oracle]

ok so you want more than 1 "bump"

[vestal fractal]

what is another school of maths to allow me to make whacky graphs

yes i dont wanna be quadratic anymore

[silk osprey]

by bumping up mean the point

here

[visual oracle]

is this somethng you want

[vestal fractal]

yeah something like that

[visual oracle]

i see

use a higher degree polynomila

[vestal fractal]

i would like to no longer be quadratic

[silk osprey]

yea it’s not a quadratic

[silk osprey]

^

[visual oracle]

this is the function for that

do you know what a root is

[vestal fractal]

i think im gonna need khan academy for this

[vestal fractal]

a type of vegitable unless its squared then its the inverse square of a number

[visual oracle]

in terms of polynomials

not like square roto

like what is a root of a polynomial

[vestal fractal]

polynomials is when the depression happened

so i think i need to learn polynomials on khan

[visual oracle]

yeah maybe do that

probably teaches better than me

[silk osprey]

💀

not quite more like the zeros of the function or where the function crosses the x axis

[vestal fractal]

so, polynomials will allow me to create more interesting and multi-variable graphs that don't just change the amplitude, phase, and angle of the function

[vestal fractal]

then i can put a new function there

[silk osprey]

well polynomial functions in this case aren’t multi variable

[vestal fractal]

and connect it to a new curve

[visual oracle]

amplitude and phase are words that describe periodic functions

not polynomials

btw

[vestal fractal]

the only math i learned by myself is trig

[silk osprey]

r u learning math in school?

[vestal fractal]

no

[silk osprey]

r u homeschooled?

[vestal fractal]

im a software engineer who suddenly needed math

[silk osprey]

oh

[vestal fractal]

i was neglected as a child and dropped out

[silk osprey]

sorry to hear that

seems like u have done well for urself if ur a software engineer

i’d watch videos on khan academy or organic chemistry tutor

[vestal fractal]

yeah I mean, i got my first job, last year

it lasted 7 months

😎

[silk osprey]

or u can buy some algebra books

[silk osprey]

unfortunate

[vestal fractal]

no i mean thats pretty good methinks

19 year old, after all

[silk osprey]

ahead of most

[vestal fractal]

anyway, Its more that, i've gotten far inspite of the fact i don't know math

so polynomials are what im looking for

not quadratic equations

[silk osprey]

yea just learn about functions

it should be algebra 2 and precalc material

[vestal fractal]

i hope its intuitive to me given my job

thank you

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[eager moon]

How would i prove this result. Note we have collection of centered random variables X1, ..., Xm such that E[Xi] = 0 for all i = 1 to m taking values in [a, b] in R, and XBar is the maxi Xi

[obsidian stream]

I don't really know how the problem will shake out when you actually do it, but I assume you apply Jensen's Inequality and that hint, then it should just all naturally fall into place without much grief

[remote arch]

Did you try out Hoeffding's lemma with $\bar{X}$ instead of $X$?

[twin meteor]

OneTrackPony

[eager moon]

trying it right now, one sec let me see what i get

so i would get e^-lE(XBar) * E(e^lXBar) <= e ^hoeffdings_quantity above

so i can take log and multiply by 1/l

so i get log E(e^l*XBar) <= E(XBar) + l * (b-a)^2 / 8 where l is lambda

so i guess i gotta find an upper bound for E(XBar) which is of the form 1/l * log(m) for some m? @remote arch

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[cerulean isle]

hi

[cerulean isle]

is this correct

[paper depot]

,rccw

[paper depot]

f(x) is not cos(e^x), i am pretty sure...

also this says "calculator active problem"

you are supposed to use a calculator

like the kind that can do integration

[full arch]

integral of sin(e^x)

[cerulean isle]

oh it’s negative yea?

[cerulean isle]

i prefer paper

[paper depot]

you cannot do this on paper

[cerulean isle]

you do it out n then put it into a calculator

[daring bolt]

Integrate from 0 to 2 then use FTC

a closed form doesn't exist for this in terms of elementary functions which is why you have to evaluate it numerically

Runge Kutta or WA

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@cerulean isle Has your question been resolved?

[cerulean isle]

huh

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[open mist]

I'm looking at eigenvectors and I'm confused how alpha= root(3)/3

I don't get why it isn't 3/root(3)

It's just the normalisation part

I can see it comes from root(1+1+1)/(1+1+1)

But I don't understand how to get to it

[hushed pewter]

3/root(3) is already greater than 1. How can that be the normalization constant?

[open mist]

how do we get to root(3)/3?

[open mist]

Why can't it be greater than 1?

[obsidian stream]

You're just finding the length of the vector [1,1,1]

[drifting jackal]

Because the magnitude is 1

The magnitude of x1 is 1

[hushed pewter]

You want alpha to be a constant such that your eigenvector has unit length

[drifting jackal]

To normalize a vector, you divide each component by the magnitude. The way the answer key wrote it was it factored out the magnitude and rationalize

[open mist]

Ohh, so why wouldn't it just be 1/root3 ?

Do it just rationalised 1/root 3 in teh answer

[hushed pewter]

That's what root(3)/3 is. Either is fine

[hushed pewter]

Yes

[open mist]

Ohhh that makes so much sense

Thank you so much

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[stuck girder]

Hi

[worthy citrus]

?

[stuck girder]

Ok

I need help

With math

[worthy citrus]

probably best to not immediately spam your channel with stickers then

[stuck girder]

Ok

[worthy citrus]

read this, post ur q, wait for help

[stuck girder]

Ok

.close

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[tepid basin]

Can someone please explain this?

[tepid basin]

.close

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[earnest portal]

Please help

[astral pilot]

c=-6

Use division algorithm

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[earnest portal]

Can you explain

[astral pilot]

Do you know Euclid's division lemma

[earnest portal]

No

[astral pilot]

Quotient*Divisor = dividend + remainder

[earnest portal]

Ohh yea I know that

I got it

Thanks

[vocal sleet]

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[tulip sleet]

How do I check when a function is injective?

[tulip sleet]

Lets say I have 2+sin(3x)

[vast shale]

when the implication [
\m f{x_1} = \m f{x_2} \Implies x_1 = x_2
]
is satisfied for all $x_1, x_2$ in the domain of $f$

[tulip sleet]

can you give a specific example

so the domain is R here

[vast shale]

easiest way is to find a counterexample to the above implication

[paper depot]

sine is a periodic function

[vast shale]

say sin(0) = sin(pi) (I'm ignoring the constants because they won't change the result)

[tulip sleet]

just a random example

[paper depot]

degrees

[vast shale]

made it radians lmfao

[vast shale]

but 0 is not equal to pi so the implication is obviously false

[tulip sleet]

yes okay sin0 and sin pi are the same

[vast shale]

yes but 0 and pi are not

[tulip sleet]

right

Actually, let me give another example, because sin is indeed a bad one

$y=\frac{e^x-5}{e^x+1}$

[twin meteor]

rainy

[tulip sleet]

🙂

domain is R again

[vast shale]

yeah again like

[
\f{e^{x_1}-5}{e^{x_1}+1} = \f{e^{x_2}-5}{e^{x_2}+1}
]

[vast shale]

simplify this and see if you get x_1 = x_2

[tulip sleet]

Okay and what does that tell me?

[vast shale]

that shows this implication

[tulip sleet]

because the question is when

not if

[vast shale]

meaning it is injective

[vast shale]

well over the entirety of its domain of course...

[tulip sleet]

I see

ill try one myself now

$\frac{x+6}{x+5}$

[twin meteor]

rainy

[tulip sleet]

domain: R/{-5}

$\frac{x_1+6}{x_1+5} = \frac{x_2+6}{x_2+5}$

[twin meteor]

rainy

[tulip sleet]

$x_1 = x_2, x_1 \neq -5, x_2 \neq -5$

[twin meteor]

rainy

[vast shale]

yes

you dont have to write x_1, x_2 not equal to -5

because you already defined your domain not to include -5

so

[tulip sleet]

so i just write injective over the domain

yh think i understand it now, thx

[vast shale]

sometimes the contrapositive of this implication might be easier to prove

[
x_1 \ne x_2 \Implies \m f{x_1} \ne \m f{x_2}
]

[vast shale]

so consider what would be most fitting in your problems

[vocal sleet]

@tulip sleet Has your question been resolved?

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[bleak prawn]

Is square root zero real?

[worthy citrus]

$\sqrt{0}$?

[twin meteor]

ΣΑC

[bleak prawn]

yeye

[worthy citrus]

That is just 0

[bleak prawn]

So if i have $\sqrt{x+5}$ can the domain of f(x) be 5?

[twin meteor]

Odie

[bleak prawn]

include 5*

[scenic ravine]

yes

[bleak prawn]

Ah oke thanks

[scenic ravine]

you mean x-5?

[worthy citrus]

You mean -5

[bleak prawn]

yeye

Thanks guys

❤️

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[bleak prawn]

And congrats on becoming mod haha

[vocal sleet]

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[regal cloud]

Let K be a field and n ∈ N. Provide a basis for HomK(K^n, K^n). Proof!
Hint: Represent each F ∈ HomK(K^n, K^n) as a matrix, and consider a suitable basis of K^n×n.

I understand what a field is and I understand Vectorspaces, subvectorspaces, but I do not understand what HomK(K^n, K^n) means and how I find a basis for that

[hard atlas]

Hom(K^n,K^n) is the space of all homomorphisms K^n -> K^n

[regal cloud]

ChatGPT explains it like this and I kinda understand what I have to do, but I am still unsure

[hard atlas]

actually correct. in general not a good idea to use chatgpt tho if you cant verify what it says

[regal cloud]

Maybe I look up some videos before solving the assignment, I come back to this if I have any more specific questions

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[fleet frost]

Can someone check that 6/sqrt(85) is correct?

[paper depot]

seems correct...

[fleet frost]

Oh

That quick?

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[spring tapir]

is the demoniator of 2s+1, i dont understand how its chosen

[iron parrot]

What is the original question?

[spring tapir]

There exists a unique positive 𝑟 ∈ ℝ such that 𝑟^2 = 2. We denote 𝑟 by √2. im not quite sure how this will help with my question lol, but here.

[hard atlas]

they give a whole extra paragraph about how it might seem strange how the expressions are chosen

[spring tapir]

I understand that, but that doesnt explain to me why theyre chosen

or whats going on

ïs it just to make this positive?

[hard atlas]

look carefully at the proof, see where the expression appears again and why it works

[spring tapir]

it comes up when you go (s+h)^2 -s^2

which im assuming is what theyre saying that h is a factor of that expression

why the subtraction of s^2???

okay i get the point of the demoniator now, but i dont get the point of subtracting s^2 from (s + h) ^2

oh wait nm, you subtract away and then bam

Gotta keep my head cool and stomach warm.

[spring tapir]

Thanks for not telling me, i get it now. That was hard to understand lol.

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[vast shale]

can i express this integral:

[vast shale]

$\int \frac{\sin(x)}{\sin(x)\cos(a) + \cos(a)\sin(x)} ,dx$

[twin meteor]

snow

[vast shale]

as

$\int \frac{\sin(x)}{\sin(x)\cos(a)} ,dx + \int \frac{\sin(x)}{\cos(a)\sin(x)} ,dx$

[twin meteor]

snow

[vast shale]

or am i not on the right track

[daring bolt]

you are not lol

[vast shale]

why

[daring bolt]

its like saying 1/(5+2)=1/5+1/2

[vast shale]

oh is it because

i dont have like

sinx + 1

on the nominator

so i can split it like

[daring bolt]

you know the trig addition formulas?

what can we write the denominator as?

[vast shale]

its okay i will figure it out

thank you tho

:)

appreciate it

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[astral pilot]

Use trig

Let one side be x

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@vast shale Has your question been resolved?

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[vast shale]

how do i start this question

[timid turret]

lol i guess you could just compute random values and you'll eventually find the one that suits the best

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

is that acc how u do it?!

[vocal sleet]

@vast shale Has your question been resolved?

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[vocal sleet]

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[strong jetty]

is this correct?

[lusty snow]

Nah

[strong jetty]

if Y = 0 and I'm trying to find x

[lusty snow]

I don't know about Y but to find x you need to

find the roots of x²-4

but, the dots will be empty

Cause it's in the denominator

[strong jetty]

I thought if I equal a fraction to a 0 then the denominator is erased

so I thought I found the dot at ( 25 , 0 )

[lusty snow]

ok, i will explain: if you have fraction it means or upper side is equal to zero or bottom is equal to zero

[wary mantle]

Not really, for x being the roots of x^2 - 4, the fraction will be indeterminate

[lusty snow]

25≠0

[strong jetty]

oh I see

[lusty snow]

ah

My brain

[strong jetty]

that means the bottom is equal to +-2

[lusty snow]

yes, but no

[strong jetty]

can some one please explain this to me

[lusty snow]

okay

no solutions i guess

[wary mantle]

Yes

[wary mantle]

Those are not the solutions

[strong jetty]

this is what I understood

[lusty snow]

You know that denominator never equals to zero

[strong jetty]

yes

[lusty snow]

if you put 2 or -2 it will be zero

[strong jetty]

but I want to know for the next time when I have more numbers

[lusty snow]

So no solutions

[strong jetty]

also X cannot equal to +- 2 in this function

[wary mantle]

When you have something of the form $\frac{f(x)}{g(x)} = 0$, you need to find the zeros of $f(x)$.

[strong jetty]

so I understand that there is no solution in this one

[wary mantle]

In this case, $f(x) = 25$. So no zeros.

[lusty snow]

i mean maybe g(x) is also necessary cause maybe similar roots?

[wary mantle]

By finding the zeros of the denominator, $g(x)$, you just determine which points to \emph{not include in your domain}.

[strong jetty]

and if g (x) = 25 instead?

[wary mantle]

Similar roots? Do you have an example? It shouldn't be necessary to check g(x)

[lusty snow]

nah

[wary mantle]

Well, say we had $\frac{x^2 - 4}{25} = 0$.

[lusty snow]

in equations where we have same x in bottom and top

[wary mantle]

Then $f(x) = x^2 - 4$, $g(x) = 25$.

And so the zeros are at $f(x) = 0$.

[wary mantle]

Ah, you mean $\frac x x$ for example. In that case, $f(x) = 0 \iff x = 0$ would be the zero of our numerator, but that is not in the domain of the function, so no zeros

You have to look at $g(x)$ to determine the domain of not already given

[strong jetty]

so this has a solution, but if we flip the fraction then we dont have a solution

[wary mantle]

Yes

[strong jetty]

thank you

.close

[vocal sleet]

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[vocal sleet]

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[empty linden]

Trig identities

[vast shale]

Where's the question?

[empty linden]

I don’t know what I’m doing wrong

Can anyone help with this proof?

[loud walrus]

Can you translate into latex? I'm not able to read that

[empty linden]

I don’t know how to use latex, but I can send the printed version

1 sec

It’s part a @loud walrus

[loud walrus]

So you have to prove this?

[empty linden]

Yeah

I then simplified it as cos(2x+x)

Then tried to solve it from there

When I did that, I got 1-4sin^2xcosx instead of 4sin^2xcosx

Wait, I have an idea

1 sec

[loud walrus]

Let me do it, one moment

[empty linden]

Nvm it didn’t work

[loud walrus]

Done

Have you tried, instead of expanding, reducing the RHS?

That's how I did

Use this

[empty linden]

Wait, that’s a thing?

[vast shale]

yes

you can prove it from sin^2(x) +cos^2(x) =1

also known as pytagorean identity

[empty linden]

Yeah but I got cosx - 2sin^2xcosx - 2sin^2xcosx

[empty linden]

I see

I’ll try using that then

Wait

On the left side?

Oh, nevermind. Right side

[loud walrus]

you should get

and you can continue from there

[empty linden]

Are there any more identities I should be aware of?

Other than these?

[empty linden]

I’ve been trying

But I’m still not getting the correct answer

I’ve been on this one question for over an hour

I might just give up and use chatgpt honestly

[loud walrus]

You can use this.

cos(A)cos(B) = 1/2[cos(A+B)+cos(A-B)]

and apply to this part

and after you do it, you finished

the exercise

[empty linden]

So it’ll be equal to cos(3x) after I do it?

[loud walrus]

after you do it, it will be a trivial sum which will be equal to cos(3x)

[empty linden]

Dude there’s no way…

@loud walrus I realized that this was literally all I had to do

I FEEL SO STUPID

I DIDNT EVEN DISTRIBUTE THE COSX PROPERLY SO IT DIDNT WORK

[loud walrus]

Did you try what I told you?

Use this cos(A)cos(B) = 1/2[cos(A+B)+cos(A-B)]

in this specific part

What's the value of A and B?

This is cos(3x) + cos(-x)

since cos(-x) = cos(x)

[empty linden]

Ok

[loud walrus]

the exercise is finished

cos(x)+cos(x)+cos(3x)-2cos(x)

[empty linden]

Then expand the sum and difference ones right? (Which are cos(a+b) and cos(a-b)

[loud walrus]

= cos(3x)

You can do both directions

but don't try both at the same time xd

[empty linden]

Alright

Thanks

[vocal sleet]

@empty linden Has your question been resolved?

[vocal sleet]

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[opal flare]

Hey guys, I'm trying to understand a specific part of a proof of why I is ideal - Ring theory

$r,s \in R$ (R is a ring) $ a\in I,~I\leq R$ a connective subgroup.

$(r+I)\cdot((s+a)+I) = (r\cdot(s+a))+I=(rs+ra)+I=(r+I)\cdot(s+I)=rs+I$

I understand why those are true but then they suggest that $rs+ra-rs=ra \in I$ and I don't understand why they went to this line and why it's true.
Any help?

[twin meteor]

meitar5674

[half imp]

ra is in I because a is in I

any multiple of something in I is in I

[worthy citrus]

they dont know that I is an ideal yet i dont think

if x + I = y + I, then that means x-y is in I. apply this with x = rs + ra and y = rs

[vocal sleet]

@opal flare Has your question been resolved?

[opal flare]

Can’t it be that x-y=0? Or 0 also in I so we are ok?

[opal flare]

Exactly

I mean, I got to the same point but couldn’t understand why can’t x be equal to y

[worthy citrus]

that could be the case but yeah then the statement still holds bc 0 in I (since I is a subgroup under addition)

[opal flare]

Oh right so it just doesn’t matter whether or not it’s 0 cuz anyway it gonna be in I

Hopefully I just got the transitions and their correctness correctly

[worthy citrus]

yeah whatever x and y are, if x + I = y + I, then x - y = something in I (this could be zero)

[opal flare]

Ok awesome

Thank you!

[vocal sleet]

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[halcyon turtle]

can anyone explain this step

[heady ibex]

where the white is? They're saying that the proportion of l, m, and n is $mⓂ️-2m$

[twin meteor]

Katharine

[heady ibex]

stupid emojis

[median crane]

where the white is? They're saying that the proportion of l, m, and n is $m:m:-2m$```

[heady ibex]

mⓂ️-2n

ffs

stop this

[twin meteor]

snow

[halcyon turtle]

yep got it

tysm

.close

[vocal sleet]

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[heady ibex]

m/Ⓜ️-2m

fuck discord

[vocal sleet]

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[vast shale]

I am not quite getting this

[vast shale]

anyone mind like, explaining this alternatively?

ok i think i get everything except what in the yellow

why does adding 6 'fix' the sum?

and also i dont get where the extra 1 is coming from when they talked about the maximum sum

like obviously the biggest number you could have in BCD is 9

so 9 + 9 = 18 so where is this 1 carry coming from

[paper depot]

im gonna say something badly worded: it's the difference between the decimal and hexadecimal coding

[vast shale]

yeah i understand that how does that actually fixes it

hmmmm

actually

before we get to that

can you please elaborate on why 10011 is the maximum sum you can get and not like

10010

[paper depot]

9+9+1 carry

from prev

[vast shale]

previous?

[paper depot]

like from the previous digit?

[paper depot]

first line in this image

[vast shale]

like wait i guess illustrating by an example is best

say you are adding 9 + 9

[paper depot]

yes then you'll get at most 18

[vast shale]

1001 + 1001 = 10010

yeah

[paper depot]

they're also considering that a 1 carry might have come in from the previous digit

since yknow you gotta do that

for multi digit addition

[vast shale]

okay yeah i get that now

[vast shale]

oh WAIT i think i totally get it now

the previous carry are those like 1's right

that came from the addition of 4+6 (for the middle 1) and 8+7 (the last 1)

thanks

.close

[vocal sleet]

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[mystic remnant]

hi

@upbeat maple

[vocal sleet]

@upbeat maple Has your question been resolved?

[runic basalt]

do you know that equation of pair of angle bisector of ax^2+2hxy+by^2 is (x^2-y^2)/(a-b)=xy/h

[scenic ravine]

wait what does this even mean?

This, makes no sense to me, what does this mean

[civic sequoia]

it's a question on equation of angle bisector of pair of lines

[scenic ravine]

ah, Ok

[civic sequoia]

just compare the given equation with the standard form and substitute

[scenic ravine]

OK, then factorise to obtain the lines and then compare to the given options

nvm, that will be nasty here, better use @civic sequoia 's idea

[civic sequoia]

idk how to use the bot , i'll just paste the formula here

[vocal sleet]

@upbeat maple Has your question been resolved?

[topaz eagle]

using formula seems quickest. even though i've never heard of this formula, but it seems right. sorry i lacked the time to derive it myself.
another way is to factorize the two given quadratic polynomials (with real-valued coefficients).
hint: ||complete the square and use difference of two squares||

[vast shale]

The answer is supposed to be 5

But I got something else

Can someone walk me through how to solve that equation

What did I do wrong

[vocal sleet]

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[chrome ridge]

in the following image

[chrome ridge]

Is AOEC a square?