help-36

[obtuse hedge]

For higher n you may be able to split it into 3

3+

I doubt it but I've only been able to try small n so it could well be

[barren pebble]

We can clearly "feel" the difference between even and odd n, so intuitively the f(n) function will have floor/ceiling function, so for large enough n where the decimals jump up a unit, mtt's bound will be off

That's my intuition

[obtuse hedge]

Yeah the original sequence I was following does

These two don't diverge until n = 14 sadly which is a lot of checking XD

I will say though this sequence corresponds to a lot of things

[barren pebble]

It would be faster to just check until mtt's bound isn't tight

But again, this

[tight geode]

i have been working on this in the past week in my math class, here is what I have

[soft zealot]

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[obtuse hedge]

This would suggest that there exists no way to do it for n = 3 in less than 3 moves, but you can do it in 1 move 🧐

Hopefully this thread is open in the morning

There's a lot of interesting things this sequence enumerates that I'd like to see if I can relate to this

Also, from the starting position in standard chess, minimum number of captures by pawns of the same color to place n of them on the same file (column). Beyond a(6), the board and number of pieces available for capture are assumed to be extended enough to accomplish this task. - Rick L. Shepherd, Sep 17 2002
For example, a(2) = 1 and one capture can produce "doubled pawns", a(3) = 2 and two captures is sufficient to produce tripled pawns, etc. (Of course other, uncounted, non-capturing pawn moves are also necessary from the starting position in order to put three or more pawns on a given file.) - Rick L. Shepherd, Sep 17 2002
Like this one

[tight geode]

wait wait, I sent my first draft and that was what ny professor asked

this is the latest version

[soft zealot]

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[tight geode]

the formula also works for 3

so generally it would be $\frac{(n-1)(n-2)}{2}$

[soft zealot]

DM ModMail for new nickname

[obtuse hedge]

That's what I posited, but

[tight geode]

is there a more optimal way or what?

[obtuse hedge]

Well for n = 5 that gives 6 but you can do it with 4

Again that bound assumes you treat the whole thing as one set but if you split it into two and reverse both subsets it takes less time

[tight geode]

hmm, yea, I think we need to respect the circular symmetry

I think we are treating them as people standing in line

[obtuse hedge]

Yes, in that case you can't split them up

But since they're in a circle you can (at least into two)

Which lowers the upper bound to I believe what's given by that sequence but I would have to evaluate properly

[barren hound]

@tight geode you had already made a bad impression due to your name. What you posted was indistinguishable from LLM slop (seriously, why would it be formated that way?). You seem to be an actual human and are participating in this discussion, so I'm not going to mute you (though I have deleted the annoyingly long messages). But I would suggest having a look at the #rules before you continue.

[tight geode]

it is because it was in my prevoius assignment, it was already formatted like that, I just copy pasted it here

[trail mango]

you’re not fooling anyone blud

[barren pebble]

🍿

[tight geode]

wait wdym, im a math major and this was in one of my previous assignments, since it was a graded work I formatted it in LaTeX and it was formally written, I just thought it would help here

[trail mango]

blud……

[strange sparrow]

blud……

[obtuse hedge]

Got the whole community in here

[tight geode]

why is everyone blud..... ing me?

[trail mango]

i’m a much more professional troll than you are. you’re not getting anything through

[barren hound]

ok whatever just get back to whatever it is that this channel is about

[obtuse hedge]

oh that's hayley

[trail mango]

female pure

[barren hound]

pure female

[barren pebble]

Idk what just happened, but anyways...

[tight geode]

so are long texts not allowed?

[quiet lagoon]

allowed

[trail mango]

chatgpt copy pastes from people who have no idea what they are talking about are not allowed

[obtuse hedge]

But yeah I actually conjecture that we can't split the circle into more sets than 2 (we can call it having a genus of 1), although I can't prove it right now

[iron frigate]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tight geode]

well ask me if you think I have no idea

[obtuse hedge]

Oh there's a generating function representation of this sequence if only I knew how to interpret those

[iron frigate]

lol

[barren pebble]

Well because we can see overllaping patterns in the "solutions" like the fast-3 and fast-4, my idea is to set up a recurring sequence

Then find the general formula

[obtuse hedge]

The recurrence is as stated

(n-1)(n-2)/2

Because you just move each number into position

One by 1

With 1 2 3 4 5 you move 1 over by 4 positions to get 2 3 4 5 1, then move 2 over by 3 positions to get 3 4 5 2 1, then move 3 over by 2 and 4 over by 1

Sum of the first k-1 positive integers to reverse the order of a row of k numbers

[barren pebble]

Is this your f(n)?

[thin cloud]

sorry, a fast-3 consist of 3 moves right?

[obtuse hedge]

What would be left is to prove that the greedy algorithm is best there

[obtuse hedge]

Yes

[obtuse hedge]

No, f(n) is lower because you can split the circle in half

If n is even, then you have

[thin cloud]

But n=4 can be done with 2 moves

[barren pebble]

Uhmm... Can you please restate it?

I didn't see any

[obtuse hedge]

How did you do it in 2 moves

[obtuse hedge]

Not for f(n) but for the fast-k sorry

Which was fast-(k+1) = fast-k + k moves

[cerulean epoch]

swap 1 and 2 then swap 3 and 4

[obtuse hedge]

Now to this

Woah how'd I miss that

[tight geode]

but moving over 1 by 4 isnt necessary, because of its cirular symmetry, it does not change anything

[obtuse hedge]

Oh the sequence does have a 2 in it

[barren pebble]

Our values so far:
f(3)=1
f(4)=2
f(5)=4
f(6)=6
f(8)=12
f(9)=16

[obtuse hedge]

I guess I just wrote it wrong

But yeah the one I believe is correct (only verified up to n = what nerdyasianguy sent) has 2

I just made an oopsies

[obtuse hedge]

This is a fast-5, not n = 5

[barren pebble]

What's the 9-move sol for n=7? There should be one

[obtuse hedge]

n = 5 we split into a fast-3 and a swap of 2

[obtuse hedge]

Yeah it's up there (far away now)

But I can describe it since I have my strategy that I still haven't gotten the chance to give you my formula for

Since everyone has questions

Which is

[barren pebble]

Okay just saw it

[obtuse hedge]

Split 7 into a fast-4 and a fast-3

[tight geode]

this series looks like it has patten, you add one to the increment every 2 numbers

[obtuse hedge]

6 + 3 = 9

[barren pebble]

I'll just try to find a pattern with those then prove it later
f(3)=1
f(4)=2
f(5)=4
f(6)=6
f(7)=9
f(8)=12
f(9)=16

[obtuse hedge]

I mean you have the pattern with those so far which was floor(n^2/4) or floor(n/2)*ceil(n/2)

But I'll give my explanation quickly now that I have a chance

Ah damnit I always write this formula wrong

Let me fix

[barren pebble]

f(5)=f(4)+f(3)+1
f(6)=f(5)+f(4)+0
f(7)=f(6)+f(5)-1
f(8)=f(7)+f(6)-3
f(9)=f(8)+f(7)-5

[obtuse hedge]

Ah it's too far up

[barren pebble]

Eh... Doesn't seem like any oattern like this

[obtuse hedge]

But the sum of the first k-1 positive integers is k(k-1)

Which is what I should have written

[barren pebble]

Hold up

[obtuse hedge]

Now using this we get mtt's bound

[barren pebble]

f(5)=f(3)+3
f(6)=f(4)+4
f(7)=f(5)+5
f(8)=f(6)+6
f(9)=f(7)+7

f(n)=f(n-2)+(n-2)

[barren pebble]

f(10)=f(8)+8=20 would match this

Yeah this seems like it

Induction time?

[obtuse hedge]

If n is even, then we split n = 2k into 2 fast-k's
That means it'll take 2*k(k-1)/2 = k(k-1) = n/2(n/2-1) = n^2/4 - n/2 = (n^2-2n)/4

If n is odd, we split n = 2k+1 into a fast-k and a fast-(k+1)
That means it'll take k(k-1)/2 + k(k+1)/2 = k/2(k-1 + k+1) = k^2 = ([n-1]/2)^2 = (n-1)^2/4 = (n^2 - 2n + 1)/4

[barren pebble]

I'm not really following the fast-k's stuff

[obtuse hedge]

You should get the same answer, try it

[thin cloud]

I also got this, idk if my argument is the same as your

[obtuse hedge]

To state it more formally for when you get back

[barren pebble]

My recurrence gives n(n-2)/4 for even n and (n-1)²/4 for odd n

[obtuse hedge]

Yes that is what I have

👌

Which is what it should be since the terms currently match the sequence

[barren pebble]

What's the argument behind it though?

Btw i'm impressed how after that many years, the question is still challenging

[thin cloud]

Where's this question from

[obtuse hedge]

The fastest way to reverse a list of $k$ elements takes $\frac{k(k-1)}{2}$ transpositions.

To work on this problem, we split the circle into as close to halves as possible:

If $n$ is even, then we split $n = 2k$ into two lists of length $k$, which we swap the order of:
That means it'll take $$2*\frac{k(k-1)}{2} = k(k-1) = \frac{n}{2}\left(\frac{n}{2}-1\right) = \frac{n^2}{4} - \frac{n}{2} = \frac{n^2-2n}{4}$$ transpositions to complete.

If $n$ is odd, we split $n = 2k+1$ into one list of length $k$ and one of length $k+1$, which we swap the order of:
That means it'll take $$\frac{k(k-1)}{2} + \frac{k(k+1)}{2} = \frac{k}{2}(k-1 + k+1) = k^2 = \left(\frac{n-1}{2}\right)^2 = \frac{(n-1)^2}{4}$$ transpositions to complete.

[barren pebble]

Compared to like IMO1986/P6 used the be one of the hardest olympiad questions but now it's commonly taught to highschoolers as an example

[barren pebble]

This is Vietnam's 1990 TST

[thin cloud]

Okay lmaooo

the 2 havles part is the same as mine

[barren pebble]

Where's the first claim from?

[soft zealot]

Coolempire93

[obtuse hedge]

By the induction I demonstrated earlier

Although that's not to say I could prove it trivially

[barren pebble]

Btw just to add, we can write f(n) more neatly with celing function because n(n-2) and (n-1)² are off by exactly 1

So f(n)= ceiling of n(n-2)/4

[obtuse hedge]

Yes, which is how they get what they wrote in the actual sequence

With this being the generating fucntion

[obtuse hedge]

I only saw it after observation, how did you come up with it?

[thin cloud]

Well I do some small cases

And it seems like

[barren pebble]

Have you proved that the halving thing is optimal?

[thin cloud]

2 sequences are being reversed in order

[obtuse hedge]

I have not, simply that it's an upper bound 👍

[thin cloud]

So I think of a bracelet

[obtuse hedge]

I do believe that I can just strum up a bijection to this situation with the chess pieces though

What's weird is the order mattering for the circle but not the pieces

[thin cloud]

It's weird tho because I wasn't able to finish the proof

[obtuse hedge]

I also think the one about the bipartite graph could be directly related to the halving idea

a(n) is also the maximal number of edges that a triangle-free graph of n vertices can have. For n = 2m, the maximum is achieved by the bipartite graph K(m, m); for n = 2m + 1, the maximum is achieved by the bipartite graph K(m, m + 1).

[thin cloud]

I thought of an actual bracelet that have numbers on it xd

kinda funny

[barren pebble]

New math merch idea

[obtuse hedge]

Interesting

[thin cloud]

I mean, like how would you actually make the order counterclockwise on a bracelet

You cut it

And then connect them the other way around

That's how I came up with the idea

[obtuse hedge]

Ah I found this recurrence in there as well

a(n) is the sum of the positive integers < n that have the opposite parity as n.
Deleting the first 0 from the sequence results in a sequence b = 0, 1, 2, 4, ... such that b(n) is sum of the positive integers <= n that have the same parity as n.

I might have to create an account just to leave a comment that this sequence is the minimum number of steps it takes to counterclockwise order n beads on a bracelet

Supposing we can prove it's optimal

[cerulean epoch]

i think if you sum the distances from where they start and where they need to go and then divide by 2, you will get something very close to this

this is a lower bound by the way

if you are odd then you will have one fixed point

[obtuse hedge]

Anyway I really do need to sleep so I'll be going but I'm going to favorite this channel and I'll be returning to this problem regardless 😆

[thin cloud]

I think split the circle into 2 halves is the only way to can make it counterclockwise ( this is where I kinda stuck) , if we can show that, we only have to show k-swap + k+1 swap is the lowest

Which shouldn't be too hard considering we already have closed form of the sequence

[thin cloud]

Gn

[barren pebble]

For each n, there can be many many solutions

The k-swap with k+1-swap is just one solution

What we need to prove is that this yields the lower bound

Note that the recurrence of mine above:
f(n)=f(n-2)+(n-2) yields the same result

So i'll just prove f(n)≥f(n-2)+(n-2)

(f(3)=1 and f(4)=2 are trivially checked by hand)

[thin cloud]

I found the solution online

[barren pebble]

If we add 2 people (n-1 and n) to the n-2 people and consider the chord formed by them in the circle

[thin cloud]

Letme see

[barren pebble]

Then we can treat swapping with n-1 or n as "jumping" through the chord

Because we need to move the n-2 original people through the chord, it takes at least n-2 moves

This is the desired lower bound

[thin cloud]

I believe we can only do 2 big-swap

[barren pebble]

The k-swap with k+1-swap above shows that this can be achieved, so QED

[barren pebble]

You could probably prove it, I don't have to just for the sake of the problem

[barren pebble]

@obtuse hedge read from here when you wake up

I think this is done

[thin cloud]

If I can show that the rest would be trivial, considering we have closed form of k-swap

I just don't have any more energy to do so, Imma head out

I have solution here if you want to see, tho I don't think it shares similar idea as our

[barren pebble]

I think the "jumping over the inaginary chord" argument above is sufficient

[barren pebble]

Nah I'm good, I know where to find the sols when i need to

Thanks anyways, bye

.close

[final saddle]

Channel closed

Closed by @barren pebble

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[final saddle]

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[maiden oracle]

i can’t figure out this problem,my basic idea is to extend the functional to the entire C([(0,1)] and use riesz representation theorem ,but to extend it I need to prove the functional is bounded,that is where I am stuck

[final saddle]

@maiden oracle Has your question been resolved?

[final saddle]

@maiden oracle Has your question been resolved?

[cerulean epoch]

,rotate

[cerulean epoch]

have you tried mimicking the proof of riesz representation theorem by hand

[maiden oracle]

to use the riesz representation thm I first have to extend the functional

[cerulean epoch]

i don't think you can do that in general

i meant like the actual proof of riesz representation where you first define μ on open sets U by taking sup of l of functions whose range is contained in [0, 1] and whose support is contained in U

blah blah blah

except you can't do this immediately since X might not have these functions so you need to modify the construction slightly

[final saddle]

@maiden oracle Has your question been resolved?

[maiden oracle]

i wanna know wether X contains the constant function one

if that is sure then we are set

[final saddle]

@maiden oracle Has your question been resolved?

[final saddle]

@maiden oracle Has your question been resolved?

[deep condor]

so you can't know which functions are in X (could be all nonconstant), but you know that X is a dense linear (Banach) subspace of C[0,1]. can you use the properties of the subspace to show that a constant function must be in X?

for gmz's solution, can you use the properties of the subspace X to show that μ exists for U? (i.e. that there are functions with range in [0,1] and support in U)?

that is the extent of what i know, so you can try #advanced-analysis if you dont get any more help here. enjoy the problem :)

[final saddle]

@maiden oracle Has your question been resolved?

[karmic shard]

[final saddle]

@maiden oracle Has your question been resolved?

[deep condor]

@maiden oracle, since you have not closed the channel yet, can you tell me if you tried the above method? where did you get stuck / what aren't you sure about?
if you do not need this channel, you may type .close to close it :) if not, please engage with the solutions or voice your concerns, otherwise this channel will be closed soon

[native cave]

i

[maiden oracle]

.close

[final saddle]

Channel closed

Closed by @maiden oracle

Use .reopen if this was a mistake.

[harsh talon]

This problem asks you to show that a positive linear functional defined on a dense subspace of C([0, 1]) can be represented by a unique Borel measure. This is a classic application of the Riesz-Markov-Kakutani Representation Theorem, but with a twist: the functional is initially defined only on a dense subspace X.
Here is the step-by-step logic to solve it.

1. Show that l is bounded (Continuous)
   Even though the problem says l is "not assumed to be continuous," a positive linear functional on a space of continuous functions is automatically bounded.

• Let f \in X such that |f|_\infty \leq 1. This implies -1 \leq f(x) \leq 1 for all x \in [0, 1].

• Since X is a dense subspace, the constant function \mathbf{1} might not be in X, but we can work around this. For any f \in X with |f|_\infty \leq 1, we have:

• By the positivity of l, if f, g \in X and f \leq g, then l(f) \leq l(g).

• This implies that for any f \in X, |l(f)| \leq C |f|_\infty. Thus, l is a bounded linear functional on the subspace X.

2. Extend l to the whole space C([0, 1])
   Since X is a dense linear subspace of the Banach space C([0, 1]) and l is a bounded linear operator, there exists a unique continuous linear extension L: C([0, 1]) \to \mathbb{R}.

• For any g \in C([0, 1]), we define L(g) = \lim_{n \to \infty} l(f_n), where {f_n} is a sequence in X converging to g.
• Crucially, this extension L preserves positivity. If g \geq 0 in C([0, 1]), we can find a sequence f_n \in X such that f_n \to g and f_n \geq 0 (or slightly shifted to ensure positivity), meaning L(g) = \lim l(f_n) \geq 0.

3. Apply the Riesz-Markov-Kakutani Theorem
   Now that we have a positive linear functional L on C([0, 1]), we apply the representation theorem:

• Theorem: For every positive linear functional L on C([0, 1]), there exists a unique Radon measure (which, on a compact metric space like [0, 1], is a Borel measure) \mu such that:

• Since L is an extension of l, it follows that for all f \in X:

[errant oasis]

What is this ai slop

[strange sparrow]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[muted prairie]

wow his account is already gone

[final saddle]

Available help channel!

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[twin pivot]

$E[X]=Np ,P(X_N = k) = \binom{N}{k}, p^k, (1-p)^{N-k}$

[twin pivot]

can someone justify this?

wait

[tired walrus]

just send an image mate

[soft zealot]

Goofy Joe

[twin pivot]

this 2

[small grail]

looks like binomial distribution

[tired walrus]

yeah looks like binomial distribution

[twin pivot]

could you help me understand what these formulas mean, please

[tired walrus]

the binomial distribution arises when you repeat the same experiment n times, where the probability of success in each trial is p (constant across all trials), the trials don't affect each other in any way, and you count up the total number of successes

[twin pivot]

Thanks

Why does PMF have that formula there?

[forest totem]

helps to try examples with small numbers

say if N = 3 and you are flipping a (biased) coin with Heads (0.6) and Tails (0.4):
if you want to find the probability of HHT, that's just 0.6 * 0.6 * 0.4
and then HTH = 0.6 * 0.4 * 0.6 which is the same
and then THH = 0.4 * 0.6 * 0.6 is the same again

so there are 3 ways you can choose 1 tail out of 3 flips

that also means, there are 3 ways you can choose 2 heads from 3 flips

(3 choose 1) = 3, (3 choose 2) = 3

you should be able to find the factorial definition of n choose r

essentially, you can swap the r heads with each other, and that doesn't give you a different order, so there are r! duplicates of the same sequence of heads and tails

and then you can swap the (n - r) tails with each other, so (n - r)! duplicates

[twin pivot]

why 3 ways

[forest totem]

you don't need the PMF to understand this

[twin pivot]

HHT

[forest totem]

HHT, HTH, THH

that's 3 ways to have only 1 T

[twin pivot]

ah so the order doesn't matter?

[forest totem]

yeah!

[twin pivot]

but couldn't we also say that for example THH there are 3 ways to arrange these 3 elements!, but for example writing THH and THH is the same thing, so I can divide by 2! and 1!, and to obtain the probability then divide by all the cases which are 8?

it seems the same to me,but I don't know if it's just a coincidence

[forest totem]

there are 2^3 = 8 cases in total

but we're only considering the cases that have only 1 T

[twin pivot]

yes

[forest totem]

actually this is where Pascal's triangle comes from

1, 3, 3, 1
0T, 1T, 2T, 3T

and the total 1 + 3 + 3 + 1 = 8

[forest totem]

I'm glad you're realising something

yes, you can also do 3! / (2! * 1!) = 3

and that's the proper way to find the number of cases that have only 1 T

[twin pivot]

yes

[forest totem]

arrange the Ts among each other; arrange the Hs

[final saddle]

@twin pivot Has your question been resolved?

[twin pivot]

thanks

[final saddle]

Channel closed

Closed by @twin pivot

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thin cloud]

Find the shortest distance between the parabola and the circle

[thin cloud]

Yeah I forgot how to do these problem

[forest totem]

it's when the curves have the same normal line

so one condition is that the tangent slopes must be the same for both points

[thin cloud]

and how does that work

[thin cloud]

I did thought of that but I'm not sure why

[forest totem]

it's a variation on, "the shortest distance between two parallel lines is the perpendicular distance"

I think it's easier to consider the case when you want to find the shortest distance between a point and a curve

[thin cloud]

but how do we know it would make perpendicular distance

Like

How do we know that line connecting those 2 points is perpendicular to the tangent

[forest totem]

you can imagine making a circle with any radius centered at that point

then there's a unique shortest distance, where the tangent slope is perpendicular to the slope connecting the point and some arbitrary point on the blue curve

[thin cloud]

the question is, why

[thin cloud]

okay hang one I'm still on this one

Why do we know that would be perpendicular

like in the image

[forest totem]

that's what we're building up to

[forest totem]

I want you to understand this diagram first

[thin cloud]

okay

[thin cloud]

this's okay

Now what

[forest totem]

right, so you don't actually need the two curves for your problem

just take the centre of the circle and that parabola

[thin cloud]

Hey that's what I did

And the answer is wrong

,w optimize sqrt((x-30)^2+(-0.3x^2+6x-50)^2)

[forest totem]

okay let's take a look

,w d/dx sqrt((x-30)^2+(-0.3x^2+6x-50)^2) = 0

[soft zealot]

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[thin cloud]

,w sqrt((x-30)^2+(-0.3x^2+6x-50)^2) when x=11.492

[soft zealot]

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Requested by yobi_fx.

[thin cloud]

this's wrong

This's what I did in the exam

[forest totem]

https://www.desmos.com/calculator/busm69g6hw

looks tangent to me

[thin cloud]

The answer was 11.7, and I'm certain that's what they give us

[forest totem]

oh lol

you forgot to subtract the radius of the circle

[thin cloud]

Ohhh

DAMN ITTTT

I THOUGHT MY SOLUTION WERE WRONG

[forest totem]

yeah and that is 11.7, I just confirmed

[thin cloud]

FUCKKKKKK

😭

Okay that method I get it, I meant I even used it

But I still don't know why

the perpendicular part here

[forest totem]

it just follows from substituting y = -0.3x^2 + 6x into sqrt((x - 30)^2 + (y - 50)^2)

I think the more important part is:

why when finding the extrema of sqrt((x - 30)^2 + (y - 50)^2)

you can just find the extrema of (x - 30)^2 + (y - 50)^2

[thin cloud]

I'm using wolfram, it's powerful enough to just give me the answer

THIS

I DON'T GET THIS

[forest totem]

oh okay

I can explain

[thin cloud]

okay

[forest totem]

you want these black lines to be the exact same line

[thin cloud]

why

[forest totem]

like, to find the shortest distance between two curves

you choose one point on the other curve, let's focus on the point and the red parabola

find the shortest distance for that point

then choose another point on the blue circle, and find the shortest distance to the red parabola again
over all points on the blue circle, choose the MINIMUM distance

but you could also choose a point on the red parabola, then find the shortest distance to the blue circle

both of these methods have to give you the same minimum distance

the minimum distance between (any point on red) and (any point on blue)

= minimum distance between (any point on blue) and (any point on red)

[thin cloud]

Okay I see

[forest totem]

so it's the same line

and that line has two right angles

[forest totem]

merge both these pictures together

[thin cloud]

okay I just lost 5% mark for that shiiiii

Damn it

I triple checked it btw

And somehow still wrong

[forest totem]

it's amazing you can deduce this from the simpler case of point and a curve

that's the bigger picture

[thin cloud]

yeah

[forest totem]

okay thank god you understood

[thin cloud]

I'm not that bad bro

Tho I should probably start grinding optimization problems atp

[forest totem]

I think that's a pretty hard type already

there's also like finding a circle tangent to three functions...

absolute hell.

[thin cloud]

I'll be fineeee

[forest totem]

I don't think there's any class that teaches this kind of optimisation

everyone just self-learns it

if you have Desmos and WA it's okay

[thin cloud]

I haven't done anything optimize problem fo 3 months now, I expected this

[forest totem]

the original problem isn't bad (single-variable)

[thin cloud]

who needs teacher

[forest totem]

but then there's all these variations which are challenging

[thin cloud]

Ah there're also a really fun optimization on the test kek

1 min I'll show you

Tbf I spent the last 3 months doin combinatorics so my calc skill is kinda rusty

[forest totem]

it's okay, I'm rusty in combi too

[thin cloud]

a dude want to go from point A to point B at the other side of the river. His swimming speed is 1 m/s, and the water's speed is also 1 m/s. He can run on land with speed of 3 m/s

Point A,B like so, the distance between both side of the river is 300 m and HB=900 m

What's the minimum time it takes him to go from A to B

[forest totem]

oh that

I've seen that before in textbooks

[thin cloud]

what book

[forest totem]

can't remember

oh, the twist is that the water has some velocity

usually the water is still

[tranquil pine]

from H to B it is 900 or 800?

[thin cloud]

900

[forest totem]

900

[small grail]

avg jee type q

[thin cloud]

ah yess, that alone killed perfect mark haha

[thin cloud]

oh really

[small grail]

yh they teach us in syllabus

[tranquil pine]

so prob he should minimize time in water or smth

[thin cloud]

hmm I wonder how you would do it

[tranquil pine]

and thats the point C

[thin cloud]

We have to figure it out ourself tho they ain't teach nothing

[small grail]

something like :
break his velocity in components then add velocity of water in horizontal direction the find time it takes to reach other shore wrt to theta then write the distance left to walk on land
add these all and find maxima wrt to theta

eh i need to rephrase into something understandable but i gtg now

[thin cloud]

working with angle is ass

There's a way to do this w/o angle tho

aight Imma dip

.close

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[exotic shale]

What way are there to find maclaurin series

Without using the formula and getting derivatives which is time consuming

[scarlet sequoia]

Do you know the maclaurin series for $\frac{1}{x+a}$

[soft zealot]

Rafilouyear2026

[exotic shale]

Well ik for 1/(1-x)

[scarlet sequoia]

Ok

[scarlet sequoia]

So how could you get it for this type of function

[exotic shale]

Which is 1+x+x^2+...

[exotic shale]

Maybe make it 1-(5x+6)/that

[scarlet sequoia]

Don't think about your exercise specifically yet

[scarlet sequoia]

If we can't answer this then we can't really move on to the exercise

[exotic shale]

I take out x^2 so i have 1/(1+5/x + 6/ x^2)?

[exotic shale]

Oh yes ik

I thiught you asked for this

[scarlet sequoia]

That's not the idea for the exercise

[scarlet sequoia]

For your og exercise, we're gonna need to rewrite this in terms of 1/(x+a) for some values of a

If you can see how, then that's a first step

[exotic shale]

Ok so fron 1/(1-x)

I multiply by -a

So i have 1/(ax - a)

[scarlet sequoia]

Ok, we're getting somewhere

[exotic shale]

And i set u= ax

So i have 1/(u-a)

[scarlet sequoia]

It's almost that, you're forgetting some constant

[exotic shale]

Wdym

[scarlet sequoia]

1/(1-x) = -a/(u-a)

Because you multiplied by -a on the denominator, you need to do the same on the numerator

[exotic shale]

So i have

a+ a^2 x + a^3 x^2?

[scarlet sequoia]

Uh

Did you mean u instead of x

Also wait

It doesn't make a lot of sense

[scarlet sequoia]

What is this supposed to be equal to

[exotic shale]

1/(x-a)?

[scarlet sequoia]

That's not it

[exotic shale]

How

Isnt it 1+ r + r^2 +....

All multiplied by a

And r = ax

[scarlet sequoia]

1/(1-r)

All multiplied by a
a/(1-r)
And r = ax
a/(1-ax)

So not 1/(x-a)

[exotic shale]

then what is it

[scarlet sequoia]

Back to this

What is 1/(1-x)

[exotic shale]

1+x+x^2....

[scarlet sequoia]

Ok

-a/(u-a) = 1+x+x²+...

So, how do we get 1/(u-a)

[exotic shale]

These the ones im supposed to know

[exotic shale]

Divide by -a

But x=u/a

[scarlet sequoia]

Yes

[scarlet sequoia]

Yes

1/(u-a) = -1/a - x/a - x²/a ...

1/(u-a) = -1/a - u/a² - u²/a³ ...

And we're done

And since this is only valid for x in (-1,1)

It corresponds to u in (-|a|,|a|) (accounting for both possible signs of a)

Does that make sense?

[exotic shale]

Idk this seems complicated

You sue its the easiest

Isnt it easier if we bring it to 1/(1-f(x))

So then its just 1+f+f^2 ..

[scarlet sequoia]

If f(x) is something like (x+1)², what do we do

The best possible case, and possibly the only case your method works, is if f(x) = constant*x

Which exactly corresponds to 1/(x-a)

[exotic shale]

Why

Cant we have x^2 aswell

We just put it in stead of r

[scarlet sequoia]

In your exercise, no

You can probably extend it to f(x) = constant*x^p

But that's as far as it goes

[scarlet sequoia]

We're doing the exact same reasoning, only the other way around, to get 1/(x-a) Maclaurin series

[exotic shale]

This is an example my prof did

Maybe you can see what method he uses

[scarlet sequoia]

Well it's maclaurin of 1/(1-u) with u = x³, sure

Because the denominator is already nice looking

[exotic shale]

Ah ok

So in my example

If i take out x^2

Its 1-(5x+6)

Then take out 5

[scarlet sequoia]

Huh

[exotic shale]

No wait

I take out 6

And have 1+5/6 x + 1/6 x^2

And i set u = 5/6 x + 1/6 x^2 and have 1/(1+u)

[scarlet sequoia]

Two things wrong

I'll start with the thing we discussed earlier

[scarlet sequoia]

You'll have a lot of problems with f not of this form

And second thing

[exotic shale]

Why

[scarlet sequoia]

u doesn't even approach 0

So you can't use maclaurin

[exotic shale]

Oh btw

It says to find maclaurin up to third order factors

I forgot to say that its important though

So maybe the approach is different

[scarlet sequoia]

Up to third order only?

It does clear up the path for a third way, closer to what you've been trying to do

But let's try to continue with what I've been hinting at the whole time

Can you factor the denominator?

[exotic shale]

Are you sure

I dont want to do something more complicated than i need for exam

The questions only ask up to 3rd or 4th or 5t

[scarlet sequoia]

If you insist

Don't divide by anything other than constants

If you divide by x² for example, the thing you did before still doesn't work

[scarlet sequoia]

Because of the second problem

So, make the 1 appear at another place

Like in here: x²+5x+6

[exotic shale]

Ok

So i make it 1+5/6 x + 1/6 x^2

And set u= -5/6 x - 1/6 x^2

So its 1+u +u^2 +...

Btw what does 3rd order factor means ?

Up to seeing x^3?

[scarlet sequoia]

All terms above x³ don't matter

[scarlet sequoia]

1/(denominator) yes

[final saddle]

@exotic shale Has your question been resolved?

[final saddle]

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[near thistle]

if i use a geometric argument and track where the basis vectors go for a 45 degree rotation clockwise i end up w $\frac{\sqrt{2}}{2} \begin{pmatrix} 1&1\-1&1 \end{pmatrix} $ but applying that same linear transformation on my graph i get a 45 degree anticlockwise rotation instead, w applying meaning subbing in $\frac{\sqrt{2}}{2} (x+y)$ for y and $\frac{\sqrt{2}}{2} (-x+y)$ for x. why is there that diff?

[soft zealot]

Cera

[near thistle]

context of this is im being asked to use standard result for an upright parabola at origin $4ay=x^2$, where a is focus and directrix displacement from origin to find cartesian eq of parabola w focus (2,2) and directrix $x+y+4=0$, and the method suggested by textbook is 'use a matrix transformation'

[soft zealot]

Cera

[near thistle]

hence why the 45 degree rotation clockwise for curve $x^2=16\sqrt{2}y$

[soft zealot]

Cera

[final saddle]

@near thistle Has your question been resolved?

[final saddle]

@near thistle Has your question been resolved?

[loud sundial]

Your matrix calculates the new position of a \textit{point} that is rotated $45^{\circ}$ clockwise. However, when applying this to a graph, you're effectively asking "What points land on the graph \textit{after} being rotated clockwise?" In other words, you're defining the pre-image of the graph, or just the original graph being rotated counter-clockwise. So to rotate the graph clockwise, you need to use the \textit{inverse} transformation instead. (You can also this as rotating the axes instead.)

[soft zealot]

Civil Service Pigeon

[near thistle]

oh right ok

it might take me a bit to rlly accept it

tysm for the explanation

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[loud sundial]

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[final saddle]

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[abstract bramble]

Find the general solution to $y^\prime + 3y = t + e^{-2t}$.
Textbook answer:
$$\mu(t) = e^{\int 3,dt} = e^{3t}.$$
Thus
$$
e^{3t}(y' + 3y) = e^{3t}(t + e^{-2t}),
$$
or
$$
(ye^{3t})' = te^{3t} + e^{t}.
$$
Integration of both sides yields
$$
ye^{3t} = \frac{t e^{3t}}{3} - \frac{e^{3t}}{9} + e^{t} + C,
$$
where integration by parts is used on the right side, with $u = t$ and $dv = e^{3t} dt$.
Division by $e^{3t}$ gives
$$
y(t) = Ce^{-3t} + \frac{t}{3} - \frac{1}{9},
$$

[soft zealot]

k.tten

[abstract bramble]

is the textbook answer missing a term in the general solution??

original from textbook

[scarlet sequoia]

What would the missing term be?

Wait a bit

Yeah the e^t shouldn't disappear when multiplying with e^(-3t) xd

$y(t) = Ce^{-3t} + e^{-2t} + \frac t3 - \frac 19$

[soft zealot]

Rafilouyear2026

[abstract bramble]

ok so i am not crazy thanks

[scarlet sequoia]

Which doesn't change the final answer but still

[abstract bramble]

.close

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[magic oracle]

How do I add exponents

[magic oracle]

@final saddle

[plucky rover]

,tex .exp rules

[soft zealot]

Xavier 🌺

[magic oracle]

I didn’t learn it yet

How do I know if this is a reliable source

[abstract bramble]

google it

[magic oracle]

google is full of liars

[abstract bramble]

alright derive it yourself then

[magic oracle]

what does derive mean

[abstract bramble]

discover it yourself

[plucky rover]

You aren't looking for political news, you're looking for math rules

[magic oracle]

mmm 🤔

How do I discover it

[willow jungle]

Guess and check

[magic oracle]

What does that mean

[willow jungle]

Mess with exponents and see what they equal

[magic oracle]

Is it like using a dictionary

Mb it slipped

[plucky rover]

<@&268886789983436800> might wanna check delete logs for this one

[magic oracle]

I had it on clip board

@golden merlin

@plucky rover help

[willow jungle]

Like 2^2 * 2^2 = 2^4

[magic oracle]

Wouldn’t it be 8 since there is 4 2s

[plucky rover]

I already sent you the list of rules, which you're apparently skeptical of

[tulip coyote]

Yeah, I really don't trust you're here in good faith, or genuinely asking questions

[magic oracle]

Can you source it

I know it didn’t just come from your butt

[plucky rover]

Yeah me neither especially after that deleted message

[tulip coyote]

You can come back after 10mins if you insult people again, your timeout will be much longer than that, if we don't ban you

.close

[final saddle]

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[thorn yew]

bro mods who did he insult bro...

i think he was genuinely being curious on where exponent rules came from

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[fair hemlock]

@golden merlin @plucky rover I need help on my math for algebra 2

[tiny gorge]

hey look, Bat and Bot

[fair hemlock]

Can u help me on these problems thx

[fair hemlock]

Yep lol

[final saddle]

Please do not ping individual helpers unprompted.

[fair hemlock]

Whoops

Plz help

[orchid osprey]

what do you need help with

[fair hemlock]

Cud u check if these answers are right?

[orchid osprey]

sure

[final saddle]

@fair hemlock Has your question been resolved?

[final saddle]

@fair hemlock Has your question been resolved?

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[brazen breach]

any1 w ideas

[brazen breach]

I think the logx is the ln thing

I tried using uhh

ibp but I ended up with the integral arctanx/x dx

I got no idea how that one works

I wonder if I can do that funny thing

oh yeah it's definitely 0, yeah nvm I got it

euler did this one apparently

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[lilac moat]

Just sub x -> 1/x

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[rustic wedge]

how do i show $x^2 = 2^{1-x}$ has exactly 1 solution?

[soft zealot]

rak³en

[rustic wedge]

my idea is to first show that there is no negative solution

and then do some growth rate arguments

the issue is

how tf do i show theres no negative solution

[rustic wedge]

and also, surely theres a better way than wat me is doing?

also, x in R

[cerulean spire]

Try applying log on both sides and try

[lilac moat]

I would think about showing that $f\left(x\right)=x^{2}-2^{1-x}$ is monotonic increasing

[soft zealot]

Roy

[rustic wedge]

oh shit why did this not come to mind

ty

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[desert mantle]

<@&268886789983436800>

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[remote spoke]

How on earth do i find ABD

[remote spoke]

im like totally lost

[final tangle]

the angle or area of the triangle

[remote spoke]

area

if i try to subtract ecb, i cant find aed

if i subtract acf, i cant find dbf

i thought that their heights were the same

so dfb was 2 * acf

but no

what do i dooooo

@final tangle

and centroid doesnt work either

[final tangle]

coordinate bash should work

[remote spoke]

wtf

[final tangle]

there could be something else

[stone wagon]

area ECB = area AEB

[remote spoke]

this is amc8 😭

im cooked

[stone wagon]

i wonder if that's helpful

let me try

[remote spoke]

hmm

so AED is 1 less than DBF

[stone wagon]

yes

or wait how did you get that

[final tangle]

Please halt aid, while we clear some stuff up

in relation to amc8

.close

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[orchid crystal]

Every subgroup of an abelian group is normal. Give an example to show that the converse is not true.

[orchid crystal]

I did the first part but can't show the converse is false

I mean non abelian is easy but then also ensuring every subgroup is normal is not something i can come up with

[final saddle]

@orchid crystal Has your question been resolved?

[hot hornet]

try some small groups

[final saddle]

@orchid crystal Has your question been resolved?

[final saddle]

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[warm python]

To start I prove $R$ is an additive group. Let $x,y \in R$. Then $v(x), v(y)≥0$ so $v(x+y)≥0$. So $x+y \in R$.We now prove that $R$ has the identity. $v(1)=v(1)+v(1)=0.$ so $1 \in R$.

[scarlet sequoia]

"Let x,y in R" is what you meant right

[warm python]

yea correcting that now, whiled adding a proof of inverses

[scarlet sequoia]

Also, if you take x,y in R, it's not always the case that they are in K^×

And x+y is not necessarily in K^× either

[soft zealot]

wai

[warm python]

0 isn't mapped to anything?

hmm

okay, got it

v(xx^{-1}) =0 \implies v(x)=-v(x^{-1})$
but $v(-xx^{-1}) =0 \implies v(x)=v(x^{-1})$. So $v( x)=0$. Thus inverses are in $R$ too, so $R$ is an abelian group

[soft zealot]

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

[scarlet sequoia]

You're trying to prove R is an additive group

So what is supposed to be the "inverse" rule?

[warm python]

if x \in k^x so is x^{-1}

[scarlet sequoia]

That's not what I said

You're saying R is an abelian group, but compared to which operation

[warm python]

right

[scarlet sequoia]

So, when you're trying to prove "inverses" are also in R

[warm python]

oh, I see the issue now

[scarlet sequoia]

Are we interested in x^{-1}?

[warm python]

no

-x

Well, I've shown if x is invertible v(x)=0 so V(-x)=0 so -x \in K^x

[scarlet sequoia]

Huh?

[warm python]

I have shown v(x)=v(x^{-1})=0 no?

[scarlet sequoia]

The proof is wrong

[warm python]

how so

[scarlet sequoia]

Why would v(-xx^-1) = 0 imply the next line

[warm python]

$v(-xx^{-1})= v(-x)=-v(x^{-1})$. And $v(-x)=-v(x)$

[soft zealot]

wai

[scarlet sequoia]

When did you prove this first set of equalities

And "v(-x) = -v(x)" is also false

I agree that v(-x) = -v(x^-1), as long as you managed to prove v(-1) = 0

[warm python]

yup, just figured

[scarlet sequoia]

But why would this all be equal to v(-xx^-1)

[warm python]

Well v(1 * -1)= v(-1)= v(1)+v(-1) so v(1)=0 . It then follows v(1)=v(-1) + v(-1). so v(-1)=0

well, so we've shown if x is invertible v(x)=0. haven't we

[scarlet sequoia]

No

Check your proof again

[warm python]

right 😭

[scarlet sequoia]

Also, "if x is invertible" means "if x is not 0"

So you're basically saying the whole valuation is always 0

Hard to believe

[warm python]

lemme work a bit

is this supposed to be an easy problem 😭

[scarlet sequoia]

You've got the idea to use -1 so yes it should be easy from now on

Just organize your proofs cleanly

And don't get caught into trying to prove something you're not asked to (and is false)
Like trying to show x^(-1) is in R when all we asked is to show -x is in R

[warm python]

yup

oh, v(-x)=v(x)+v(-1)=v(x)

so we're done

The next part of the problem asks me to show that if x is a unit, v(x)=0, so I thought this would kill 2 birds with one stone

[scarlet sequoia]

Well, we're done for the additive group part

[warm python]

Lemme prove it's closed under multiplication

[scarlet sequoia]

Also, you still haven't fixed those

[scarlet sequoia]

When x isn't 0, yes

Separate case for x = 0, even if it's trivial

[warm python]

they're in R

[scarlet sequoia]

Ok, but are you allowed to take their valuation

[warm python]

wdym

[scarlet sequoia]

"Let x,y in R. Then v(x) >= 0 and v(y) >= 0"

Is that always true

[warm python]

well, the set R is those elements of K^x such that v(x)≥0

[scarlet sequoia]

Reread the definition of R

[warm python]

$R= {x \in K^{\times}: v(x)≥0} \cup {0}$

[scarlet sequoia]

Yes

[soft zealot]

wai

[scarlet sequoia]

So if you take any x in R

[warm python]

yes

[scarlet sequoia]

Can you always say that v(x) >= 0

[warm python]

well, it could be 0

[scarlet sequoia]

Exactly

Huh

[warm python]

oh

sorry

right

got it

[scarlet sequoia]

0 has no valuation, so we can't talk about v(x) unless we know for sure x ≠ 0

[warm python]

right

[scarlet sequoia]

So this has to be fixed

[warm python]

so I add if x,y≠0

[scarlet sequoia]

That solves part of the problem

You'll have to make a separate case when x = 0 or y = 0, but we're on the right track

Now, "so v(x+y) >= 0"

[warm python]

when x=0, v(y+0)=v(y), which is ≥0

and similarly when x=0

but when both are 0, we run into problems

well, the function isn't defined when x=0=y

[scarlet sequoia]

What's their sum if both are 0

[warm python]

so That's a non-issue

[warm python]

0, but v(0) isn't defined

[scarlet sequoia]

But do we need v(0) to know if 0 is in R

[warm python]

no, we already know 0 \in R

it's given

[scarlet sequoia]

Yep

We also didn't need to check v(x+y) whether x=0, y=0 or both

x+y = x or y, both being in R

[warm python]

I think we're done? proving closure under multiplication is a direct conseqeunce of (i)

[scarlet sequoia]

Well, as always...

[warm python]

I should show it

yes

let v(x)≥0,v(y)≥0 then v(xy)=v(x)+v(y)≥0

[scarlet sequoia]

Also

[warm python]

yes

[scarlet sequoia]

In the addition closure before

when x≠0 and y≠ 0

[scarlet sequoia]

Do you see the problem with this

[warm python]

not really

what am I missing

[spring haven]

it's the same problem you had before

[warm python]

v(0) not being defined?

[scarlet sequoia]

Yes

It's even stated in the property that you can only use it if x+y ≠ 0

[warm python]

mhm

[warm python]

so I have to come up with a work around

I'll eat and finish this if it's fine?

[spring haven]

why would it not be fine

it's your life

[final saddle]

@warm python Has your question been resolved?

[warm python]

Hmm, can i have a hint

Oh got it

V(1+0)>min(0, v(0))

So v(0) is less than equal to 0

[warm python]

Isn't v(0) not defined in the first place

Like 0 isn't even the domain of v

Wait

So just x+-x=0 and 0 is in it

😭

How did i miss that

[final saddle]

@warm python Has your question been resolved?

Channel closed

Closed by @warm python

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[rustic wedge]

god help me

[rustic wedge]

1m^3 of neon gas (C_v =3R/2) at 273.2K and 10 atm initial pressure is expanded reversibly and isothermally to a final pressure of 1 atm. find the work done

now my module reports the answer as 232.85 kJ

whereas I am getting 2332.9 kJ

which is an apprx 10 times answer

can someone verify which answers right

gpt says mines but i can trust

[vital crag]

!show

[final saddle]

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

[rustic wedge]

well its not very difficult since its T=const, we apply the ideal gas eqn to obtain that V2 = 10 m^3 or 10^4L

then w = nRT lnV2/V1 = PV ln V2/V1 = 10^4 * 2.303 * 101.3 * 10^(-3) kJ

numerically i havent made an error, so clearly either i put values wrong, formula wrong or my modules wrong

[vital crag]

and show the module's answer

[rustic wedge]

wait nvm

my modules calc went wrong

i just put in the raw numbers they put in a calc

how do i have better calc than them they use calculators for this stuff🙏

.close

[final saddle]

Channel closed

Closed by @rustic wedge

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[left trail]

how can number 2 be possible. I graphed up a scenario and there are many values of b where the equation wont be 0

[vital crag]

"at most one point" means 0 or 1

[left trail]

oops didnt read it closely . For the proving itself I was thinking of saying that f'(x) = x^3-3x+b. and then I would need to show that f(-1) = f(1) but then you have -b = b. But that seems to lend itself to there never being a point

[scarlet sequoia]

That's not really what's happening

[left trail]

ok. This was just me trying to apply rolle theorem to the problem

[scarlet sequoia]

Yeah, but you won't apply it to some antiderivative of x³-3x+b

You'll apply it to something else
||for example, reasoning by absurd||

[left trail]

if we want x^3-3x+b = 0 woudln't that mean that is f'(c)

[vital crag]

f(x) is x^3-3x+b, not f'

[left trail]

how did you see that?

[scarlet sequoia]

I mean, reasoning by absurd is pretty common

If you have to prove that a function has at most 1 zero

What happens when it has 2 or more should be impossible

And two zeroes is exactly where Rolle intervenes

[left trail]

idk what that is

[scarlet sequoia]

To prove a statement with "absurd reasoning"

Suppose that the statement is false

And find a contradiction

[left trail]

but doesn't rolle take care of 1 or more zeros

[scarlet sequoia]

You're mixing up with the derivative of the function

And it's the consequence of Rolle, not the conditions to apply it

Rolle says:
IF a function f continuous on ..., differentiable on ... and has f(a) = f(b), two points with the same image
THEN f' has a zero in between

So if you want to apply Rolle, it's no use trying to find the zeros of a derivative

Because that's what Rolle is gonna give you, not what it requires

So, you should look for a function to which you can apply Rolle, meaning a function that has f(a) = f(b)

[left trail]

I have tried that with x^3-3x+b but then you get -1+3+b=1-3+b=0. but no b exists

[scarlet sequoia]

Who said "a" and "b" were gonna be -1 and 1

[left trail]

that is the bounds of our function though

[scarlet sequoia]

Yeah, but the actual points with same image can be anywhere inside the bounds

[left trail]

doesn't the way he define it imply the endpoints?

[scarlet sequoia]

Look, in our exercise

The function f(x) = x³-3x + b has a derivative everywhere on R

So no matter which segment [a,"b"] you're looking at

f is gonna be continuous on that segment and will have a derivative on the open interval inside

When the exercise tells you to look for zeros inside [-1,1], you're tempted to take [a,"b"] = [-1,1]

But, as you said it, it won't work because f(-1) is not f(1)

So

[left trail]

ok I see. so all I need to do is find a sub interval in -1 to 1 and then make sure that there is only one x

[proven elk]

there is one thing i think you might be missing

f(a) != f(b) but Rolle's theorem is about f(a) = f(b)

in this case you have f'(a) = f'(b)

right?

[scarlet sequoia]

Ok...? But what's the point applying Rolle to f'

[proven elk]

or is it the other way

[scarlet sequoia]

You won't get anything out of it

[proven elk]

is there some g such that g'(x) = f(x) and g(a) = g(b)

[scarlet sequoia]

Still not

We're gonna apply Rolle to f itself, not "g", not f'

[proven elk]

well rolle's theorem tells us the derivative of f obtains zero

[scarlet sequoia]

Just not on [-1,1]

[proven elk]

but the question is about whether f obtains zero

that's different

[scarlet sequoia]

I think you still haven't understood the point

The point is to show f' has a zero

Because f' has no zero on (-1,1)

[proven elk]

so if you want to conclude f obtains zero at some point by rolle's you're actually talking about f being the derivative of the function that has g(a) = g(b)

why?

rolle's concludes 0 = f'(somewhere) not 0 = f(somewhere)

[scarlet sequoia]

We're not trying to show f has at least one zero... we're trying to show f has at most one zero

[scarlet sequoia]

YES

So

If we find that, when f has two or more zeros

f' has a zero

And on the other hand, f' has no zero

Then that's absurd

[scarlet sequoia]

So this assumption is wrong

And f has at most one zero

[proven elk]

does rolle's let us conclude the existence of two or more critical points?

[scarlet sequoia]

?

No, it gives the existence of a critical point of f

Critical point which cannot exist

Because f'(x) = 3(x²-1)

Which doesn't cancel on (-1,1)

[left trail]

I am understanding this correctly? For rolle we need f(a) = f(c). But they want us to show that there is at most one point where this is true. Meaning a or c but we assume there is two.

[scarlet sequoia]

Yes, suppose there is a < c in [-1,1] such that f(a) = f(c) = 0

That's the opposite statement of "f has at most one zero in [-1,1]"

[proven elk]

i can work out that rolles says "if f' has no zero in the interval and f is differentiable and continuous on the interval then f(a) != f(b)"

[left trail]

then I would need to show that rolle leads to a contraction right?

[proven elk]

i think the quoted phrase i used does it

it shows uniqueness of the zero when it exists

because our f' does have no zero in the interval

this is just converse law application

a -> b therefore not b -> not a

[scarlet sequoia]

So which values of a and b are you applying this to

[proven elk]

any zero and any other proposed zero

part of a uniqueness proof

[scarlet sequoia]

To apply rolle you need a < b

So you have to suppose there exist zeroes a < b, thus reasoning by absurd

[proven elk]

in that case you need to use f(a) = 0 in one case and f(b) = 0 in the other

in general we want to show that if a != b then f(a) != f(b) i.e. the function is injective on the interval

the converse on rolle's gives us exactly what we need for that conclusion

[scarlet sequoia]

I feel like you're overcomplicating trying to prove by "Rolle converse" by all means, when the reasoning by absurd seems easier

[proven elk]

Here's that for clarity:

Rolle's: If f is continuous on [a, b] and differentiable on (a, b) and has f(a) = f(b), then f' has a zero in (a, b).
Converse: If f' has no zero in (a, b) and f is continuous on [a, b] and differentiable on (a, b), then f(a) != f(b).

[left trail]

@scarlet sequoia the only issue I can see with rolle is that on our interval f' has two zeros and you can only have f(a) = f(c) if c,a = +/- sqrt(2), or 0 . the sqrt(2)'s are out of bounds but 0 is in bounds which ends up going against what we want since we can find a point in [-1,1] where f(a) = f(c)

[scarlet sequoia]

Where did the sqrt2 come from

[proven elk]

Those zeroes aren't in (a, b) so you're good

[left trail]

nvm I messed up with system of equations

but still we have 0 which is true

[scarlet sequoia]

The zeroes of f' are ±1, which are outside of (-1,1)

So they're outside of (a,c) where a, c are the two zeroes of f we picked

Contradicting Rolle

[proven elk]

when a or c is a zero right

[scarlet sequoia]

When both are zeroes of f (reasoning by absurd)

[proven elk]

this shows the other cannot be a zero

[left trail]

ok I see all of this now but wasn't the only way to set them equal was by making a and c 0 which means that there could only be one point to begin with

[proven elk]

it lets you conclude that they must come from equal inputs

two different inputs can produce a zero

we have shown that two zeros must be the same zero

they cannot be different on this interval

you can rearrange the contradiction proof to obtain a converse proof that involves showing f is injective on the interval

which lets you conclude the uniqueness of the zero directly as a special case

[left trail]

ok so we start with a and c and assume rolle. Then we obtain -1 and 1 for our zeros of f'. But we don't know that it is outside of [a,c] until we know that a = c = 0 and that we would have to do by directly.

What I dont' see is how we can say that -1 and 1 are not in the interval of [a,c] without knowing what a and c are which would show us that they are the same

[proven elk]

Then we obtain -1 and 1 for our zeros of f'.
Then we use the fact that -1 and 1 are the only zeros in [a, b] and neither is in (a, b)

[left trail]

where does rolle theorem say that. it only says that the derivtives must be defined at each of the poitns in (a,b) not be (a,b)

[proven elk]

ok so we start with a and c and assume rolle
We start with c being some zero and then consider [a, c] and [c, b]

[proven elk]

then f' has a zero in (a, b).

To contradict rolle's you must show that f' has no such zero

[left trail]

why?

[proven elk]

Because we are trying to prove this is the only zero, we need to check two other possible intervals that could contain a distinct zero.

We have to contradict Rolle's for both.

Once you've shown there's no zero of f' in either (a, c) or (c, b) then Rolle's (by contradiction) lets you show every d in either [a, c) or (c, b] must have f(c) != f(d) because c != d.

[left trail]

how can we have more than one zero on [-1,1]?

[proven elk]

The objective at that point is to show the opposite.

Any zero on [-1, 1] of f is unique.

[left trail]

but do use rolle we have to assume the opposite

[scarlet sequoia]

ifndr suggested using the converse of Rolle

[proven elk]

to use rolle you have to prove f' contains no zeros and then you know that f(a) != f(c)

[scarlet sequoia]

If you want to use direct Rolle, you can do the absurd reasoning we did earlier

[left trail]

but it doesn't work

[proven elk]

i am trying to think how to make it work because it should

you can just convert converse proofs to contradiction proofs

but the prose is escaping me

[scarlet sequoia]

Isn't that the point of reasoning by absurd

showing that something can't work

by assuming it does work

[proven elk]

let's say there are two zeroes of f i.e. c != d and f(c) = f(d) = 0

[left trail]

wait wait I am now getting confused

yes this I understand. Then our objective is to show that f'(x) = 0 but x not in (c,d)

[proven elk]

Rolle's: If f is continuous on [a, b] and differentiable on (a, b) and has f(a) = f(b), then f' has a zero in (a, b).

Then f' has a zero in (a, c] or [c, b), contradicting f'(d) != 0

||There i found the confusion||

[left trail]

why maybe d is in one of the intervals

[proven elk]

the premise "there are two distinct zeros in the interval" is said "suppose c and d are unequal zeroes"

and the case analysis goes and says "d must be either above or below c"

then no matter which we can use rolles to contradict that

there i got u the contradiction version instead of the converse version

had to think about it for a little while

[left trail]

I still don;t see how we show this

[left trail]

.

[proven elk]

not quite accurate there i'll help fix that up