help-33

[cosmic raptor]

but why is that?

[red nimbus]

if it has roots it changes monotony which breaks the one to one

[cosmic raptor]

but in the case of x^3, wouldn't 0 be a root?

for the derivative?

[red nimbus]

ye... i worded it wrong

i meant it must not intersect the axis

like go through

[cosmic raptor]

so it must be non-negative?

[red nimbus]

or non-positive

[cosmic raptor]

ohh okay, makes sense

thanks!

[red nimbus]

but rn this isnt making sense to me

[cosmic raptor]

so in those cases where they ask for one-to-one functions

just make sure that it is f'(x) >= 0 ?

[red nimbus]

for all t yes or f'(t) <= 0 for all t

[cosmic raptor]

wait, sorry, what do you mean for all t?

[red nimbus]

what does it mean?

we wanna make sure it never crosses the axis

which means for all values of t

we make sure it never happens

[cosmic raptor]

oh okay, makes sense

thanks so much! lifesaver

./close

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[marsh citrus]

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[red nimbus]

did you get it?

[cosmic raptor]

yup yup, makes sense

[red nimbus]

what was it

the second case seems like a contradiction

[valid wing]

what even is sech 💀

[cosmic raptor]

i assumed it like this:

Let's say a function is even, then we just take a point where it is always increasing

but if it is odd, then we can only do >0?

[red nimbus]

huh

We got first solution (-oo,2/3]

the other doesnt seem to have some

thats what i asked

[cosmic raptor]

wait, maybe i interpreted it wrong lol

wait idek

.reopen

[marsh citrus]

✅

[cosmic raptor]

isn't this just like x(t)'s range has to be in the domain of y(t)?

[valid wing]

did you get the answer to the initial question?

[cosmic raptor]

yup yup, i just don't know when to use f'(x) >= 0 vs f'(x) > 0

[cosmic raptor]

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[marsh citrus]

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[uncut venture]

Hi there I don’t know how to estimate the values of a and b on my graph? My graph was a sketch so I have no idea what to do unless I have to like correctly and accurately draw it?

[onyx flume]

what is the equation

[uncut venture]

5b

Would I just say like ln100 or should I know or can I do it algebraically?? I’m so lost

<@&286206848099549185>

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[worn wadi]

hello, can someone give me counter example or a proof that int(A) in int(B) does not imply A in B if A and B both subsets of R^n?

[worn wadi]

thnak u

[valid cape]

int?

[quaint elm]

interior

[valid cape]

oh okay

[quaint elm]

{a \in A : there is an open ball around a that is a subset of A}

[worn wadi]

yes

[hollow glen]

can you construct a non empty set, such that its interior is empty?

[worn wadi]

yes

if there's only a single element in the set?

[hollow glen]

yeah

does that maybe help find a counterexample?

[worn wadi]

uh yes thank u very much 👍

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[gray iron]

Im very stuck at a

Tbh not really sure how to start aside from writing out the circle formulas

[marsh citrus]

@gray iron Has your question been resolved?

[gray iron]

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[red skiff]

am I tripping or the this answer is wrong

isn't the last terms supposed to be 48?

[marsh citrus]

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[red skiff]

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[crude nimbus]

@hollow glen idk why the last channel closed but to continue our conversation from there, when putting A into any matrix calculator it will say the only eigenvectors are (1, 0), and (0, 1), and not something like $(x_1, x_2)$ (every vector). Are these calculators wrong?

[elfin berry]

Ezlanding

[proper pivot]

I'm struggling with intersects

[crude nimbus]

this channel isn't open rn, post this message in an open channel in Math Help (Avalible) please or see #❓how-to-get-help

[crude nimbus]

^

[hollow glen]

not wrong per se, they only list those vectors that can form a basis for all eigenvectors

[crude nimbus]

So a more correct answer for the eigenvectors of A would be every vector (something like $(x_1, x_2)$, since they are both free) , but only the basis for the eigenvectors are shown in these calculators

[elfin berry]

Ezlanding

[hollow glen]

yeah, all except for (0, 0), if you look up a general definition for eigenvectors it makes sense too

[crude nimbus]

The calculators should say something like "eigenvalue bases vectors", then, to make it less confusing :|

but yeah makes sense now, thanks for the help

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[quasi gust]

I dont know, how to derive through the solution

[quasi gust]

the solution is so complicated

im talking about 1b

[stark sandal]

is it?

[quasi gust]

I'm lost

[stark sandal]

what part do you not understand

well

!status

[marsh citrus]

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

[quasi gust]

From the beginning

[stark sandal]

they used the given relation to find $\sin(\theta)$

[elfin berry]

Percy

[stark sandal]

you can also find $\cos(\theta)$ or whatever ratio you're comfortable with, you just need one

[elfin berry]

Percy

[stark sandal]

then they made a triangle to find the tan

[quasi gust]

ok, but like, what's the point of re-writing, tetha's value

and also, if sin is more than 0, wouldn't the A and S quadrant be appropriate

[stark sandal]

if you can find a trig ratio of theta in another way go for it ;-;

this is just the most straight forward

[stark sandal]

i dont know that notation but yes it's either 1st quadrant or 2nd

you'd also need cosine value to be sure lmao

[quasi gust]

ok, then why did the answer pick the 2nd quadrant, not the first?

[stark sandal]

it's a bad solution

think of it like

you see sin of x and cos y are positive

so x is in either 1st or 2nd and y in either 4th or 1st

[quasi gust]

okay

[stark sandal]

this is a shit way of complaining im sorry

a sec lemme write stuff down myself so i can explain clearly

okay

so

x and y are acute angles

[quasi gust]

yes

[stark sandal]

$x \in (0, \frac{\pi}{2}), y \in (0, \frac{\pi}{2})$
$\implies x+y \in (0, \pi)$

this is clear yeah?

[elfin berry]

Percy

[quasi gust]

ok, that looks confusing

[stark sandal]

not studied ranges?

basically if they're both acute the maximum their sum can be is 180

and minimum 0

that's all

[quasi gust]

ok, understood

[stark sandal]

if you add another $\frac{\pi}{2}$, that range instead changes to $(\frac{\pi}{2}, \frac{3\pi}{2})$

[elfin berry]

Percy

[stark sandal]

so it can be 2nd or 3rd quadrant, actually.

[quasi gust]

oh ok

[stark sandal]

and we have $\sin(\text{\emph{this stuff}})$>0, so it must be in 2nd quadrant.

[elfin berry]

Percy

[quasi gust]

yup

[stark sandal]

hence then $\tan(\theta)$ would be $\in (-\infty, 0)$

[stark sandal]

yeah so that

[quasi gust]

oh

[elfin berry]

Percy

[quasi gust]

ok

thanks

[stark sandal]

mmm

woohoo we did it

yay

[quasi gust]

yup

thx

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[gray iron]

I know there's more solutions but idk how to find them

[lucid zenith]

cos(x)=cos(y) can also be true when x=-y, for example

[stark sandal]

hollow knight fan?

[lucid zenith]

There are infinitely many solutions like this

[gray iron]

Yes

[lucid zenith]

Not quite

[gray iron]

How so?

[lucid zenith]

What you have written are indeed solutions but you're missing a few

[stark sandal]

you have a term of $2n\pi \pm x$

[lucid zenith]

No, that's for when it's sin(x)=sin(y)

[stark sandal]

wait is this cos

right sorry

i think it's this

[elfin berry]

Percy

[lucid zenith]

$\cos(y)=\cos(x)\iff y=2n\pi \pm x$

[elfin berry]

kheerii

[stark sandal]

yeah

,rcw

[gray iron]

Sorry for rotation

These other 2 solutions look correct

[stark sandal]

just use this and you should be good

[lucid zenith]

This doesn't look right

,w cos(2x)=cos(x-pi/4)

[gray iron]

Why not?

[lucid zenith]

Yeah your solutions are wrong

The last 2

[lucid zenith]

Just use this

And solve the + and - cases separately

[gray iron]

Ahh ok

1 sec

I got the same answer

Hmm

Wait

No I didnt

Why can't I get x = π/12 + k*2π with this method

Nvm I got everything right

Thanks for the help

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[noble axle]

How do I prove this?

Don't know if there's an easy method to do this? I'm starting the chapter on Quadratics, so I would prefer not using the quadratic equation for this

[stark sandal]

well they don't use the quadratic equation now do they

[noble axle]

For some reason, I thought the book did not have a solution to the problem. I don't have the solutions set for this book. I will read the solution and get back to this and see if I need to close the thread or not.

I'm not sure if I understand the first sencond paragraph of the solution, starting with "Instead". Why would it be important for the quadratic not to equal 0, and making a big deal out of integer roots?

Also, I kind of get the if 3|n and 2|n. then 6|n part. Though, it's useful for this particular question, I would like to know why it seems to work in general.

[brave marsh]

It is true that this approach wouldn't work for non-integer roots, but at least it helps you find them, if there are any. (And since the degree of the polynomial is 2, then there can be at most two roots, so even finding one is helpful.)

Do note though that in the initial question you posted, they explicitly say that the roots are integers, so that you can search right away for those.

As for why the quadratic is not equal to 0, what they explain is essentially that if x is odd, then the whole expression is odd. Since 0 is even, then the expression can't be 0 (so x is not a root). That means that if you're looking for roots, they should be even.

[patent sandal]

If u want the proof, its as follows. if gcd(a,b)=1, a|n and b|n then n=ak and n =bm. ax + by = 1 (by bezout's theorem) multiplying both sides by n, we get anx + bny = n, then a(bm)x + b(ak)y = n
=> ab(mx+ky) = n. Thus ab|n

If u want an intuitive understanding, since 3 and 2 dont have any common factors, the prime factorisation of n must contain both 3 and 2

thus 3*2 = 6 must divide n

[noble axle]

I see.

[noble axle]

I don't get it sorry. odd numbers - even numbers would be odd. But why is it important that the quadratic has no roots at the value of 0? Maybe you answered the question, and I failed to read it.

[brave marsh]

A root of the quadratic happens whenever the quadratic is 0 (i.e. p is a root iff f(p) = 0)

That's the definition of a root

They show that if x is odd, then the quadratic expression f(p) is odd, and so it is not 0, so x mustn't be a root

[noble axle]

Is it just the fact that if the solutions of a quadratic happen to be like (x)(x-c) then there wouldn't be two roots?

[brave marsh]

No. They are not arguing anything about the value of the root other than it being even or oddd

It`s the value of the function that can't be 0 if the input is odd. By definition that input is not a root of the function in that case

All they are saying is : if the input p is odd, then f(p) is odd.
Since 0 is even, then f(p) is not 0
Therefore, p is not a root if it is odd

[noble axle]

x being a root looks like this, right? (x)(x-c)

[brave marsh]

I'm to be blamed as well because I didn't make this clear, but you're confusing the root and x

If p is a root, then it looks like f(x)= (x-p)(x-c)

[noble axle]

I'm following

Do you mind if you go through your explanation step by step, and I say "I'm following" until we reach the conclusion?

[brave marsh]

A number p is a root of a function f if f(p) = 0

[noble axle]

I'm following

[brave marsh]

For the quadratic function f you are given, if the input p is odd, then the output f(p) is also odd

[brave marsh]

p is unrelated to this here

[noble axle]

I'm following

[brave marsh]

Therefore f(p) is not equal to 0 if p is odd

[noble axle]

I'm following f(p) can't be odd when the input is odd

[brave marsh]

f(p) can't be even

you mean

[noble axle]

oh sorry, yes.

I meant that, wrote it wrong

[brave marsh]

Good

So yeah f(p) can't be even, so in particular f(p) is not 0

Because 0 is even

[noble axle]

I'm following

[brave marsh]

Since f(p) != 0, then p is not a root.

[noble axle]

Since the output does not equal 0 when given an odd number, the odd number cannot be a root?

That's how I'm interpreting it.. Probably not what you mean.

[brave marsh]

That's right

[noble axle]

I'm following.

[brave marsh]

That's it. If p is odd, then it is not a root, so you must look for even roots

[noble axle]

Oh I get it, thank you very much.

@brave marsh@patent sandal@stark sandal I am pretty bad at getting taught. Thanks for the patience, and help.

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[obtuse crown]

can someone help me with 3)b

its basically "prove that for any x of D, B(x)=..."

thanks

[fervent yarrow]

use 2 + 2cos(2x-π/4) = 2 + √2(cos2x + sin2x) (proven in 2)a))

[obtuse crown]

alright im gonna try that thanks for the idea

[marsh citrus]

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[summer cave]

Can someone help pls ?

[quiet anvil]

As mentioned in a previous help thread. The integral does not have a closed form in its indefinite form, and the definite integral is 4 K(1/2)^2.

The helpee is seeking a way of solving this.

I only know the above from using a CAS

[wary kite]

yea i know i’m just laughing because it’s always funny when someone opens a help channel, gets an answer, acknowledges it, then just opens another one later as if it hadn’t already been answered

[quiet anvil]

No, that's not it.

She already had the answer in a slightly different form. She wanted to know the process to arrive at it.

[wary kite]

ohh

[quiet anvil]

Previous thread: #help-1 message

[marsh citrus]

@summer cave Has your question been resolved?

[main idol]

yea posting without all the previous work is such a lame move

[summer cave]

He already said everything useful

But here it is

And the result should be lemniscate constant squared x 2

Btw the lemniscate constant is defined by two times the integral from 0 to 1 of 1/sqrt(1-x^4) dx

[quiet anvil]

Yes, but the point is I shouldn't have to be the one to. It implies that you're not really interested in putting in work into this problem yourself, and you're expecting someone to do the leg work for you...

This is your problem, you should be the one explaining it. No?

[summer cave]

No sorry, i really want to solve it 🙏🏻

[quiet anvil]

oh, this is quite interesting then.

what if we integrate the area by letting y = kx and integrating over x and k instead?

[summer cave]

Let’s try

[quiet anvil]

anyway, I need to go afk, so best of luck with this problem. It seems interesting so I might give it a try when I get back to my computer.

[summer cave]

Thx for ur help, take care 🙏🏻

If you find something interesting, tell me

[marsh citrus]

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[marsh citrus]

@crystal plank Has your question been resolved?

[marsh citrus]

@crystal plank Has your question been resolved?

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[stark glacier]

Hello.

I have a Statistics question about choosing a type of statistical inference for a given project.

I made a choice, but the choice might not fit with the data collected and I was wondering if I could get hints into what type of analysis would work best here.

It's about comparing two brands of popcorn. Measuring the frequency of popping kernels per time interval (measuring how many kernels pop between 0 to 10 seconds, 10 to 20 seconds, etc).I placed an audio recorder next to a microwave and created a data table with frequency and cumulative frequency of the popping kernels. A few variables were kept constant (such as method of heating, power applied, grams of popcorn), I did this 3 times for both brands, so 6 raw data tables.

The choice of statistical inference: I thought I could apply logistic regression, where the probability that a kernel has popped or unpopped is dependent on time, but it looks like I would have to do that for the approx 300 kernels that has popped so far, and for the 3 trials unless I average them out, but I don't know where that will get me. I also thought about linear regression but the rate at which the kernel will pop will be slow in the beginning, fast in the middle, and slow in the end, so perhaps a non-linear regression might be better?

I feel like I am stuck at a roadblock. I have a thick book containing fundamental statistics and inference statistics, but I can't seem to link most examples to this project. Struggling to get creative in the realm of statistics. Anyone have any insight?

[marsh citrus]

@stark glacier Has your question been resolved?

[marsh citrus]

@stark glacier Has your question been resolved?

[stark glacier]

Perhaps I'm all alone in this

[proud ice]

Probably (accidental pun )

Most helpers here don't do stats

[stark glacier]

I feel like I need a mentor then

[proud ice]

a data table with frequency and cumulative frequency
What? What are the dimensions here exactly?

[stark glacier]

the table is 2D Idk what you mean, the frequency is just how many kernels were popped per time interval.

while cumulative just sums it all up until the last time interval

perhaps i dont know the definition of dimension here

i also how many kernels were left unpopped at the end, as well as how many were wasted/badly-popped/useless. Wait--- I can maybe use that instead of time??

but--- uhh-- no

I'm so lost aaaaaaaaaaaa

[stark glacier]

.close

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[novel juniper]

Prove that $\gcd(a,b)= a \alpha + b \beta$ for some values of $\alpha, \beta$

[elfin berry]

A dense set(Ping when reply)

[novel juniper]

so first $gcd(a,b) \mid \alpha a + \beta b$ for all values of $\alpha ,\beta$, trivially

[elfin berry]

A dense set(Ping when reply)

[novel juniper]

for the other direction , I was thinking of using the division lemma

$gcd(a,b) = q(a\alpha + b \beta)+r$
\
$gcd(a,b) -q(a \alpha + b\beta) =r$
\
$0 \leq r <a \alpha + b\beta$

[elfin berry]

A dense set(Ping when reply)

[novel juniper]

I'm trying to use the well ordering principle here, not sure how to

<@&286206848099549185>

[cedar stone]

do you know euclidean algorithm?

[novel juniper]

yes

That's what I've used here

[novel juniper]

here

[cedar stone]

dyk that if you can represnt a as a linear combn of b and c, and you can represnt c as a linear combn of b and d, than you can reprsent a as a linear combn of b and d?

as in

[novel juniper]

I haven't proven that

[cedar stone]

if a = bv1+cv2
and c= bv3+dv4
than
a=bv5+dv6 for some v5 and v6

ok, try to, that is a step in the proof that ik of

[novel juniper]

Instead

I can try to arrive a contraidction, no?

[cedar stone]

yes

[cedar stone]

q is what?

[novel juniper]

quotient

let a\lapha + b\beta be the smalest element in the set

positive element anyway

[cedar stone]

you wont get a contradiction from that

[novel juniper]

hmm

[cedar stone]

instead a*alpha + b*beta, try the smallest positive value a*alpha + b*beta can take

[novel juniper]

$d= a \alpha + b \beta$

[elfin berry]

A dense set(Ping when reply)

[novel juniper]

we consider the smallest posible d

[cedar stone]

yes

[novel juniper]

so now what

[cedar stone]

so we know that gcd(a,b) divides d, right?

[novel juniper]

yes, but we want to pove the opposite direction now

[cedar stone]

yes

so try doing that

[novel juniper]

ah

so gcd(a,b)= q(a\alpha + b \beta)+r

[cedar stone]

yk if a|b a<=b (assuming a and b positive)

[novel juniper]

yes

[cedar stone]

so q will have to be 0, right?

or q =1 r = 0

[novel juniper]

what

[cedar stone]

gcd(a,b) divides d

so gcd(a,b) <= d

[novel juniper]

yes

okay

wait

got it

oh

nvm

my bad

I thought I could write gcd(a,b)= ab/lcm(a,b)

[cedar stone]

assume k*gcd(a,b) = d

[novel juniper]

Or maybe I consider $A={a \alpha + b \beta \mid a\alphaa + b \beta >0}$

[elfin berry]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel juniper]

okay

[cedar stone]

actually

an easier method would be

[novel juniper]

Actually this looks easier

[cedar stone]

yes

[novel juniper]

Got it

😭

Thanks

.close

[marsh citrus]

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[marsh citrus]

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[wispy rover]

can someone make a super long and complex polar equation in desmos that represents ssomething

[main idol]

Just make your own

[proud arch]

Based response

[marsh citrus]

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[marsh citrus]

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[faint lintel]

I am confused about It is a common mistake to choose ....

Could someone explain that for me please

[marsh citrus]

@faint lintel Has your question been resolved?

[marsh citrus]

@faint lintel Has your question been resolved?

[marsh citrus]

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[wary bluff]

Im kinda confused about this....when using the formula from image 1 what did they do with a_i ?

[wary bluff]

oh this was 5.13 btw

this was for 5.13 so for this question wont a1,a2,a3 be similiar but replace c_i with 1

[wary bluff]

this is the part im tripping abt

oops im an idiot

.close

[marsh citrus]

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[wary bluff]

wait no

.reopen

[marsh citrus]

✅

[wary bluff]

im dumb that shouldnt work either

https://tenor.com/view/sad-monkey-monkey-crying-sad-monkey-crying-gif-26326543

this is the next question (related to these 2 aswell).

and this is tripping me too

how did they transform second image formula to first image formula

<@&286206848099549185>

bruh I just realized it made use of this formula (image 1) but this was never covered in my class...is there a way to derive this formula with just this info (image 2)

[marsh citrus]

@wary bluff Has your question been resolved?

[wary bluff]

https://tenor.com/view/sussa-gif-5432297967573209335

[marsh citrus]

@wary bluff Has your question been resolved?

[wary bluff]

<@&286206848099549185>

[marsh citrus]

@wary bluff Has your question been resolved?

[marsh citrus]

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[still temple]

guus can i get anotherh elp

[still temple]

https://cdn.discordapp.com/attachments/1095362988132806707/1310923164502724618/IMG20241126190017.jpg?ex=6746fbd6&is=6745aa56&hm=7c3a48dcf4124137b27196593abffbdf1b358202e74782b72aacb1c44df2dd2e&

this is the last one bro i saewar

om jiust studying fir my math test

i tried to combine like terms

[still temple]

3y - 2y = -5 + 15

[tawdry rampart]

correct

[still temple]

so is it 1 = 10

[tawdry rampart]

that is extremely wrong

[still temple]

WHAT

1Y = 10

[tawdry rampart]

yup correct

[faint narwhal]

The two angle are not necessarily the same

[tawdry rampart]

they are

[faint narwhal]

[tawdry rampart]

angle subtended by a chord on the circumference lmao

mb adwan

[still temple]

sp uh

ooh

so 1y = 20

[tawdry rampart]

yes it should be sorry

[still temple]

oh what so its 20

then

whys the answer C here

[tawdry rampart]

they asked PSQ

not y

lol

[still temple]

but

oh uh

am i supposed to do this

3(20) - 5

okk got itt

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

UH

CAN SOMEONE HELP M

.CLOSE

.close

[marsh citrus]

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[stray fern]

Yes

[marsh citrus]

@small stream Has your question been resolved?

[marsh citrus]

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[still temple]

What do they mean by "conservative" here ?

[shadow warren]

we consider the vector field F as conservative if the line integral over any two points is the same for any path we take

[vast magnet]

bio checks out

[still temple]

so we are defining the term "conservative"

[shadow warren]

yeah essentially

[still temple]

okay

thanks

[shadow warren]

also it connects to physics if you think about conservative vs non conservative forces

[still temple]

oh yeah

.solved

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

[still temple]

how can the integral of a scalar product be the difference of 2 vectors?

oh

i thought there was a gradient before the f

mb

.close

[marsh citrus]

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[sudden badger]

im not sure how to continue part c

[tawdry rampart]

4a + 6a = ?

[sudden badger]

10a

[tawdry rampart]

similarly

[sudden badger]

10logb5?

[tawdry rampart]

correctt

[sudden badger]

oh i thought the rule for adding logs was
logaX+logaY = logaXY
and
XlogaY = logaY^X
so it would be
logb5^4+logb5^6
logb(5^4)(5^6)

[amber birch]

unfortunately 25 = 5^2 and 125 = 5^3

[tawdry rampart]

it is

[amber birch]

but yeah I see where you are at now

okay so you have $10 \log_b 5 = 5$ as stated previously

[elfin berry]

southlander!

[sudden badger]

but why isnt it logb(5^4)(5^6) as the rule suggests?

[tawdry rampart]

$\log_b{5^4}+\log_b{5^6} = \log_b{(5^4)(5^6)} = \log_b{5^{10}} = 10\log_b{5}$

[amber birch]

I think you're confusing yourself

you have the right answer

but it would be much easier to do $\log_b (5^2) + \log_b (5^3) = \log_b 5^{2 + 3} = \log_b 5^5$ first

[elfin berry]

southlander!

George (Wumpus Man)

[amber birch]

$= 5 \log_b 5$ and now you can multiply this by 2

[elfin berry]

southlander!

[tawdry rampart]

do you see what ahppened here?

instead of all that we could have just normally added

lol

[sudden badger]

oh i understand now

[sudden badger]

ah i see why that'd be easier

[sudden badger]

oh so if the end is the same we can just add it like ab+cb

[tawdry rampart]

yess

but if you let the log thing to be a
it will just be 4a + 6a which is 10a

this is what i recommend

always let the log things
solve for a then solve for log

[sudden badger]

ah i see

so b=25

[tawdry rampart]

correct

nice

[sudden badger]

wb part d

is it a simultaneous equation?

[tawdry rampart]

it is

[sudden badger]

i forgot how to do those

[tawdry rampart]

firstly ^

[sudden badger]

wait so do i get rid of the number before it?

and make it into one log?

[tawdry rampart]

that works

[sudden badger]

do i have to do change of base then

[tawdry rampart]

if the bases are different

then you cant do this

[sudden badger]

but cant i just apply the change of base rule

[tawdry rampart]

hint: ||a = log_2(x), b = log_3(y)||

||then solve for a and b||

[sudden badger]

but there isnt an a or b in the equation

[tawdry rampart]

if the question was 3a + 4b = 10, a - 2b = 1

would you be able to solve it?

[sudden badger]

yes

simultaneously

2a+2b=9

?

[tawdry rampart]

i dont think that is correct

[sudden badger]

is it a = 12/5

then sub it back in for b

[marsh citrus]

@sudden badger Has your question been resolved?

[marsh citrus]

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[novel juniper]

I'm trying to prove that the derivative of an even function is odd

[novel juniper]

so $f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$

[elfin berry]

A dense set(Ping when reply)

[novel juniper]

$f'(-a) = \lim_{h\to 0} \frac{f(-a+h)-f(-a)}{h= \frac{f(a-h)- f(a)}{h}$

[elfin berry]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel juniper]

$f'(-a) = \lim_{h \to 0} \frac{f(-a+h) - f(-a)}{h} = \lim_{h \to 0} \frac{f(a-h) - f(a)}{h}$

[elfin berry]

A dense set(Ping when reply)

[distant peak]

let u = -h
you have (f(a+u)-f(a))/(-u) = -f'(a)

[lucid zenith]

the derivative of an even differentiable function

[novel juniper]

I'm lost

sorry

[devout mauve]

||chain rule||

[novel juniper]

That works, yes

but

First principles

[devout mauve]

then do what herels said

[novel juniper]

ah

[devout mauve]

f(a-h)=f(a+u) and u goes to zero

[novel juniper]

got it

thanks

[nimble prairie]

It’s practically giving away the answer

[novel juniper]

Why does u have to be -h , though

[distant peak]

i mean, i want to get the expression of f'(a) again

[novel juniper]

yes, but limf(x+h) =lim(x-h)

[distant peak]

when h goes to 0, yea sure

[devout mauve]

its not like the last step is strictly necessary

[novel juniper]

which it is

so how's the derivative odd

[nimble prairie]

So what’s the problem?

[devout mauve]

but if say f(x) is increasing near a and h is positive, then f(a-h)-f(a) is negative, while f(a+h)-f(a) is positive

so the order does matter

[novel juniper]

ah right

the left hand limit and right hand limits will be the negatives of one another

[lucid zenith]

No

[novel juniper]

no, that's not what I meant

[nimble prairie]

I mean you know the function by assumption is differentiable, so it shouldn’t matter

[novel juniper]

I mean if the LHL for f(x) is a, then for f(-x) it would be -a

oh

nvm

got it

sorry

[lucid zenith]

You can make a more general statement about an even function which doesn't require differentiability, just for the left and right derivatives to exist

the left derivative at x=a is negative of the right derivative at x=-a

And vice versa

[novel juniper]

Yeah

got it

thanks

.close

[marsh citrus]

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[cunning oak]

suppose a,b,c are rational, and x1,x2,x3 are roots of P(x) = x^3 + ax^2 + bx + c.
prove that if x1/x2 is rational and not equal to 0 and -1, then x1,x2,x3 are all rational

[cunning oak]

i have no clue

[still temple]

x1:x2 = rational?

[cunning oak]

yeah

[still temple]

Okay

[cunning oak]

i wonder why they put not equal to 0 or -1 though

[tawdry rampart]

P(x) should be x^3 + ax^2 + bx + c

if x_1 / x_2 = 0, then x_1 = 0
so now x_2 can be irrational like root(2)
x_3 can be - root(3)
and a, b, c and will still be rational

[cunning oak]

woops

sorry

[cunning oak]

oh i understand

[tawdry rampart]

similarly if x_1 / x_2 = -1
x_1 + x_2 = 0
and they can be root(2) and -root(2)
x_3 will be Q
and a, b, c are still rational

[cunning oak]

i see

how would i prove this though

[tawdry rampart]

i dont know lol

[sturdy flower]

could maybe do something with Vieta's formulas

https://en.wikipedia.org/wiki/Vieta's_formulas

[rotund violet]

im thinking about this

im not even convinced this is true lowkey

[cunning oak]

im still confused lol😭

[rotund violet]

this is a pretty hard problem

[cunning oak]

its probably true, i got this from my regionals II qualifier for the thailand mo

[hollow glen]

we know that P(x) = (x-x1)(x-x2)(x-x3), so
-a = x1+x2+x3
b = x1 x2+x2 x3+x1 x3
-c = x1 x2 x3
but from here, idk if it's just trying out all kinds of algebraic manipulations or if there is some clever trick

[rotund violet]

oh lol i know who wrote this then

[sturdy flower]

this is along the lines of what I was thinking

[cunning oak]

?

[cunning oak]

oh i see

[rotund violet]

notably i am looking at the polynomial (x-x_1)(x-x_2)

as we can take x_1/x_2+x_2/x_1 and know that this is rational

and this is equivalent to (x_1^2+x_2^2)/(x_1x_2)

[cunning oak]

ab/c = cyc (x1 + x2)/x3

[rotund violet]

ok i mean i know the solution is "solveable" actually

as worse case, we write x_3 in terms of the other 4 variables

[rotund violet]

oh ok i have a non-vieta proof for this i believe

ok so suppose that x_1 is irrational

this implies that both x_1+x_2 and x_1x_2 are irrational

since x_1x_2x_3 is rational, this implies that x_3 is irrational

therefore, if x_1 is irrational, this implies that x_1,x_2,x_3 are all irrational

now we use the algebraic nature of the roots

[cunning oak]

what does this mean? sorry english isnt my first language

[rotund violet]

i mean that any root of a polynomial with integer coefficients can be expressed as p+q*sqrt(r), where p and q are rational

oh wait im trolling

we cannot assume this is a quadratic

[cunning oak]

o

[rotund violet]

okay let us try vieta bash

so we have cx_2=x_1

thus (c+1)x_2+x_3 is rational

thus x_3=a-(c+1)x_2, where a is the sum of the roots

we now have that x_2^2x_3 is also rational

thus x_2^2(a-(c+1)x_2) is rational

additionally we have x_1x_2+x_1x_3+x_2x_3 is rational

this is equivalent to cx_2^2+(c+1)x_3

ok so i solved this without vietas

im sure this time

let me restart

so we have that x_1,cx_1, and x_3 are roots where c is rational

the fundamental theorem of algebra implies we can factor the cubic into a quadratic and a linear term, both of which would have rational coefficients

take the quadratic in this factorization

there are 3 cases:

its either of the form (x-x_1)(x-cx_1),(x-x_1)(x-x_3), or (x-cx_1)(x-x_3)

meanwhile the linear term will have rational coefficients, implying that at least one of the roots are rational

we now prove that if one of the roots are rational that the rest will also have to be rational

we use proof by contradiction to prove this

suppose that all the roots are irrational

edit: suppose that exactly one of the roots are rational

[cunning oak]

oh

do i have to prove that the roots are real

though

[rotund violet]

we can deal with the non-real case quite easily

if the roots aren't non-real, then 2 of the roots will be complex, while the third will be real

since none of the resultant fractions of roots would be real numbers, i don't believe "rationality" is well-defined on the complex numbers

hence x_1/x_2 being rational wouldn't be well-defined, meaning that we can assume that all of the roots are real

ok, so now we assume that exactly one or two of the roots are rational

suppose this root is x_1 or cx_1.

this implies that both x_1 and cx_1 are rational, and since the sum of all the roots are rational this implies that x_3 is rational

now suppose that the rational root is x_3

in this case, we have that (c+1)x_1+x_3 is rational

since x_3 is rational, so is (c+1)x_1, implying that x_1 and cx_1 are rational

thus, all the roots are rational no matter what

if one of them is rational

and since we know that one of the roots are rational at least

this implies all roots are rational

this should solve the problem i believe

[cunning oak]

cool

im gonna read it for a bit to understand😭

[hollow glen]

why rational coefficients?

[rotund violet]

this happens to be true

a proper proof of this fact would require commutative algebra however it is commonly just assumed

specifically commutative algebra states that the field of rational polynomials is a unique factorization domain

[hollow glen]

like, if we take x³-2, how would you factor it to have rational coefficients?

[rotund violet]

sorry, we need to assume that all the roots are real for this to be true

let me verify this fact, i asked my roomate and he said it was true lol

he was an imo gold so i trusted him

[hollow glen]

even then, the linear term would still have cbrt(2)

[rotund violet]

ok so x^3-2 wouldn't work, as in our conditions we have that all 3 of the roots are real

in x^3-2, only 1 root is real

the other roots are the cube root of 2 times the third roots of unity

[hollow glen]

oh where was the condition they are all real?

[cunning oak]

where is this from?

[rotund violet]

yes

[hollow glen]

idk I dont see it stated anywhere

[rotund violet]

well first we can agree that we can factor any polynomial into quadratics and linear terms

[hollow glen]

real quadratics and linears

[rotund violet]

https://en.wikipedia.org/wiki/Unique_factorization_domain

we first prove that the set of rational numbers are a unique factorization domain

[hollow glen]

thats trivial for fields

[rotund violet]

this should do it

[hollow glen]

yeah that was uhh gauss

[rotund violet]

again i just assumed this fact on olys for like 2 years until i just came into college and actually proved it myself

[hollow glen]

but then, not sure where this brings us

[rotund violet]

this shows that Q[x] is a unique factorization domain

we take the set of irreducible elements in Q[x]

these are quadratics and linear terms by the fundamental theorem of algebra

[hollow glen]

no the fundamental theorem doesn't apply for rationals like that

it's for reals

its true for R[X]

[marsh citrus]

@cunning oak Has your question been resolved?

[rotund violet]

ok let me check again with my roomate lol

notably my Q[x] i defined doesn't assume that all the roots are real

oh oh ok

i get it

we have that f(x) and f(kx) are both in Q[x]

this implies that they will have a nontrivial gcd in Q[x]

since they share at least one root

[marsh citrus]

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[forest grove]

How do I get the roots?

[forest grove]

I used AI to translate it, so the wording is wonky, but I think it´s understandable, if you need help understanding just ask me

[rotund violet]

um this AI translation is horrible

doesn't make sense at all

[wary kite]

sum: -b/a, product : c/a

ax^2+bx+c

[mellow kraken]

This is a badly worded question. It's asking

1. To find the "roots" of the polynomial, you just solve it for x in terms of k.
2. Then, fully solve the roots knowing the sum of them is 10.
3. Then, determine which interval the product of the two roots lie on. (inferred from ambiguity in the question)

[forest grove]

Which of the following intervals should be part of the product of the roots of the equation, if you know that the sum of the roots is equal to 10?

Is this clearer?

I tried to translate it

[forest grove]

Is this clearer?

This was the original.

[forest grove]

How do you get the sum and the product, or are these just values I should memorize?

[wary kite]

quadratic formula

[rotund violet]

you should still memorize them, a more general form of this is Vieta's formulas (check the wikipedia page for more details)

[wary kite]

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

[elfin berry]

knief

[wary kite]

the two roots are the + and the -

add them you’ll get -b/a

multiply them you’ll get c/a

[forest grove]

Is it B?

[forest grove]

I got 13,33... from the multiplication so it should be B right?

[forest grove]

Is that right?

As the sum I got -3/k=10 so K=-3/10

From there I went for the multiplication, c/a, (-4)/(-3/10) and that got me 13,33...

[wary kite]

mhm then c/a = 40/3

[forest grove]

Which is yeah 40/3

So since that only fits within )13. 14( which is B, that´s the answer right?

[forest grove]

?

[wary kite]

yea

[forest grove]

Thanks

You´re the best

.close

[marsh citrus]

Channel closed

Closed by @forest grove

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[marsh citrus]

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[long bay]

the problem is that a cylinder with a radius of 7.5 cm is filled with water and when a sphere is put in the water level rises by 5cm and I have to find the radius of the sphere. I got 5,9cm but it isn’t on my answer sheet and I don’t know what I did wrong

[tawdry rampart]

show your work

then we can spot the mistake (if any)

[long bay]

One of these is the answer

Ok wait a sec

I kind of thought that the water level rose by 5 so if the sphere was a Zylinder it would have to be 5cm in height and then I took the radius and calculated volume of smaller cylinder and then I put the answer in the sphere radius finding calculation

[tawdry rampart]

im getting the same answer

maybe a problem with the question

[long bay]

I think so too

Thank you very much I will talk to my teacher about it

[tawdry rampart]

are you sure we have to find the radius?

and not the diameter

because 11.9 is an option

[long bay]

oh Woodside

[tawdry rampart]

lmao

[long bay]

Woopsie

I can’t read German

[tawdry rampart]

samee

lol

[long bay]

well thank you I feel very dumb now lmao

close.

.close

[marsh citrus]

Channel closed

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[tawdry rampart]

!done

[marsh citrus]

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

[tawdry rampart]

:)

[marsh citrus]

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[fickle shell]

Wait

I just had an idea

.close

[marsh citrus]

Channel closed

Closed by @fickle shell

Use .reopen if this was a mistake.

[fickle shell]

.reopen

[marsh citrus]

✅

[fickle shell]

Nevermind 😭

[young barn]

try evaluating the one-sided limits

[fickle shell]

What does that mean sorry

[red nimbus]

Do you know $\lim_{x \to a^+} f(x)$ right sided limit, $\lim_{x \to a^-} f(x)$ left sided limit?

[elfin berry]

𝔸dωn𝓲²s

[fickle shell]

That means that right sided limits go to infinity and left sided go to negative infinity, right?

[red nimbus]

no

[fickle shell]

Oh

[red nimbus]

let me draw something rq

[jovial osprey]

your suppose to rearrange the inequality to sove for a

actualy i could be wrong

just ignore what i said

[fickle shell]

Me when I say anything 😝

[red nimbus]

you basically approach x = a from one side respectively

meaning from the right you walk towards it

and then you consider what happens if you start from the left and walk towards it

[fickle shell]

So if youre coming FROM the right its the right limit, yes?

[red nimbus]

yes

[fickle shell]

Ah cool

I got it now

[red nimbus]

they might differ

this is why

if they don't differ then we basically say the limit exists, or also the limit from both sides exist

you usually do that when you have piecewise functions where you transfer from one function to the other

[fickle shell]

Differ from what sorry?

[red nimbus]

as you might tell they can be together, or have a jump

[red nimbus]

be not the same

[fickle shell]

Ahh yeye

[red nimbus]

likehere from left and right you dont end up at the same adress basicallyl ol

[fickle shell]

Wym by address sry?

Like just the fact that its a jump discont?

[red nimbus]

i was just trying to find an intutitive explanation

[red nimbus]

like imagine the function represents a road

you would end up somewhere else from the left than if you walked from the right to x=a

[limber hearth]

That is philosophical

I like this

[red nimbus]

basically they wont meet haha

[fickle shell]

Damn thats kinda sad 😭

But good example hehe

I get it

[red nimbus]

so you could find an a where they might meet :)

or not

[limber hearth]

Give that purple a ladder

[red nimbus]

or blue jumps down

[limber hearth]

Always go higher

[fickle shell]

To be with his friend

Yay

Happy ending

So to the actual algebra

I was thinking I set 5x^2 - 6x - 1 = 2x + a

I got 2 values for a

[fickle shell]

These 2

Which were probably not what theyre looking for

So im guessing thats wrong

[red nimbus]

Ok so it's asking for what number a it has a jump discontinuity

[red nimbus]

so actually a situation like this

the correct way to do that would be to see what is f(2)

5(2)²-6(2)-1?

[fickle shell]

Wait why

[red nimbus]

basically we wanna know what value it will be from the left

and based on that find a

since we have a parabola it will be continuous so that the limit as x->2^- will be equal to f(2) anyway

[fickle shell]

Like where the left limit is?

[fickle shell]

Where the purple guy is standing?

[red nimbus]

yes

[fickle shell]

Ok

And then we set 2x + a =/= that

[red nimbus]

not so fast

[fickle shell]

Hi chartbit!! ❤️

[fickle shell]

But thats what we want no?

[red nimbus]

yea so we get f(2) = 7

[fickle shell]

Cause if we set it =/= to that, that is all the values where theres jump discont, no?

[red nimbus]

we also know that the limit from the left of the parabola is 7 as x goes to 2

[red nimbus]

i will get there

now you wanna make sure in order to have jump discontinuity that the right side limit does not approach f(2) = 7

we work with limits to make it more rigorous

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

you wanna make sure the right side limit is not 7

[fickle shell]

So were subbing in 2 into 2x+a for the same reason as here, right?

Except from the right this time

We wanna know where the blue guy is standing

[red nimbus]

we wanna know from the blue guy's perspective how to choose a so that it cannot get to his purple friend

his purple friend is located at f(2) = 7 height

now you wanna make sure their y-levels differ

the strange thing is that there are infinite many possibilities for a

we can say let blue near 8 or 9 or -100

just make sure not 7

😅

[fickle shell]

Oke I think I get it

2(2) + a =/= 7

4 + a =/= 7

a =/= 3

Right?

[red nimbus]

yea

[fickle shell]

🎉 🎉 🎉 🎉

[fickle shell]

Thank you so much!!!

These questions used to be so easy haha

But im getting back into it slowly 😅

Thanks again

❤️

.close

[marsh citrus]

Channel closed

Closed by @fickle shell

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[red nimbus]

tbh i just explained it too much i could have said yes but i dont wanna

[fickle shell]

Noooo you didnt

I appreciate that you are teaching me the finer details

[red nimbus]

you will get it as you progress in math that your life was of lies haha

[fickle shell]

You are an amazing teacher

❤️

[marsh citrus]

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[left moat]

did i go about thhis problem correctly?

[marsh citrus]

@left moat Has your question been resolved?

[left moat]

.close

[marsh citrus]

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[marsh citrus]

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[cunning fractal]

Winnie keeps flipping a coin until she gets either two heads in a row or three tails in a row, at which point she stops. What is the probability her final flip is heads?

[cunning fractal]

I know I can approach this recursively but I'm not sure how to do so, and resources online are jarring

[latent coral]

not sure how to do it recursively, but my first guess is that you want to find the conditional probability P(last flip heads | she stopped)

[cunning fractal]

Yeah

[latent coral]

which is P(last flip heads and she stopped) / P(she stopped)

the two events for which she stops are disjoint so P(she stopped) = P(two heads in a row) + P(three tails in a row)

right?

[cunning fractal]

Yes

[latent coral]

i'm sure you know how to find those two probabilities?

so you can find P(she stopped)

[cunning fractal]

1/8 + 1/4?

[latent coral]

yeah

now considering P(last flip heads and she stopped)

she only stops in the two events: she got two heads in a row, or she got three tails in a row

the intersection of this with her last flip being heads is just the event that she got two heads in a row

[cunning fractal]

so 3/8 * 1/4?

[latent coral]

wait why are you multiplying

are you trying to find P(she stopped) rn?

[cunning fractal]

No

The P(Last flip heads and she stopped)

Is that wrong

[latent coral]

how did you get 3/8 * 1/4?

[cunning fractal]

P(stopped) = 3/8
P(last heads) can only occur if P(heads twice) = 1/4
P(both) = 3/8 * 1/4

[latent coral]

ah

P(A and B) is only equal to P(A)P(B) if the events are independent

[cunning fractal]

Ohh wait yeah I forgot that

[latent coral]

but you don't know that (last flip heads) and (she stopped) are independent

well think of it, the event that she stopped is just the union of two events

(two heads in a row) U (three tails in a row)

this is the event that she stopped

[cunning fractal]

Yeah, which is 3/8

[latent coral]

right, but you want (last flip heads) intersect (she stopped)

[cunning fractal]

Yeah

[latent coral]

so whatever events are in this intersection must also be in the event (she stopped)

[cunning fractal]

Ohh

[latent coral]

and only one event is in this intersection, and that's the event (two heads in a row)

[cunning fractal]

Wait nvm

I'm still not sure. You're using Baye's formula right?

[latent coral]

i don't think i'm using bayes theorem, i'm pretty sure i'm just using the definition of conditional probability

all i'm doing is finding P(A|B) as P(A and B) / P(B)

[cunning fractal]

Okay

[latent coral]

let me reiterate, you want to find P(last flip heads | she stopped)

which is P(last flip heads and she stopped) / P(she stopped)

[cunning fractal]

Yes

[latent coral]

you found P(she stopped) to be 3/8 since P(she stopped) = P( (two heads in a row) U (three tails in a row)) = P(two heads in a row) + P(three tails in a row) = 1/4 + 1/8 = 3/8

[cunning fractal]

Yep

[latent coral]

now you want to find P( (last flip heads) intersect (she stopped) )

to get the numerator

she only stops when one of two events occurs: she flips two heads in a row, or she flips three tails in a row

if you intersect this with the event that her last flip is heads, what event do you get?

do you see why P( (last flip heads) intersect (she stopped) ) is the same as P( two heads in a row )?

[cunning fractal]

Ohh wait yes of course

[latent coral]

great, and you already know P(two heads in a row)

[cunning fractal]

1/4

[latent coral]

which means you can find that P(last flip heads | she stopped) = (1/4) / (3/8) = 2/3