Mathematics - 2017-04-30 (page 1 of 5)

Danu

12:00 AM

you forgot z itself ;)

thanks for that tip

arctic tern

avoided using a full basis in my answer anyway

Danu

Maybe also define z in that answer

arctic tern

just edited that

Danu

(I can do it too if you'd like)

oh okay

arctic tern

I am using the fact that $a\times b=ab$ when $a\perp b$ in case I should write that in too.

Danu

12:10 AM

yeah

well it's fine by me

your rep is 9696

arctic tern

my # of answers on anon account is 1337

haha

hi chat

o/

Oh wait, has DHMO been... removed?

sorry if I'm slow at noticing things

Akiva Weinberger

12:43 AM

@BalarkaSen @Astyx Right, yeah, and other degeneracies like two disjoint triangles connected by an edge (in addition to the stuff you mentioned)

In any case, yeah, it is easily proven.

Simply Beautiful Art

Why doesn't math.SE chat rooms run Mathjax/Chatjax automatically?

Akiva Weinberger

Also, I think there's a "dual": the closure of increasing unions of triangles (or polygons of bounded amounts of sides) should also be a triangle (polygon).

I wonder if, for any set of closed sets that is closed under decreasing intersections, it's also closed under closures-of-increasing-unions.

(Or if the converse might be true as well.)

Holy shit, Turkey blocked Wikipedia?!

Semiclassical

1:43 AM

hi chat

that's...yikes

Little Rookie

2:38 AM

@BAYMAX so whats the exact reason for
If b≠a then LHS ≠ RHS ? I just want to be sure.

Akiva Weinberger

2:53 AM

We need more symbols

Let $ɯ=3$

$ð=\cos(ɯ)$

$ɫ=O(1)$

$\{ɕ:ɕ\in ʃ\}$

$Д=Я-И$

Rubertos

Seriously downvoting for no reason must be banned

Little Rookie

3:11 AM

@BAYMAX and why $b\neq a$ then the quadrature rule is not exact

Anonymous

@Rubertos Everyone has a reason to downvote but you may not like the reasons :P

Akiva Weinberger

@blue I looked at the question in question, it seemed fine

(I can words)

Anonymous

@AkivaWeinberger That's why I said that everyone has a reason, but you may not like it. Some people just enjoy downvoting!

Anonymous

I know a few people whose votes are mostly downvotes

Anonymous

Some people downvote as a revenge.

Anonymous

3:20 AM

Some people for weird reasons :P

Akiva Weinberger

I'm curious what an MSE revenge story would sound like, now

Semiclassical

pretty banal unless you forbid 'petty revenge.'

Akiva Weinberger

I need an old-school detective noire

Anonymous

@AkivaWeinberger lol, like having a quarrel with someone on mse and downvoting all of their new posts :P

Anonymous

I've seen such cases in Physics SE

Anonymous

3:28 AM

(There is a vote reversal script that runs everyday, but that's not so effective)

BAYMAX

4:48 AM

@LittleRookie I am not getting how you got that two expression you had shown earlier!,like I tried putting trial simple polynomials like $x^0,x^1,x^2,..$ until we donot get same values (for both quadrature rule and the integration value)and when that happens say for $x^{(n+1)}$ then we say degree of precision of that Quadrature rule is $n$.Is this good! or any other approach you are using!?

I would also like to see why the limits of integration matters while calculating the degree of precision of the Quadrature rule?

Eric Stucky

@TheGreatDuck: Hey, are you online?

TheGreatDuck

@EricStucky what a nice coincedence

your definition of the implied integral has become obsolete.

Eric Stucky

hehe

TheGreatDuck

5

Q: Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me. Let us define a di...

Eric Stucky

Is the 'implied derivative' that you're using on your newest question the same as what you were trying to do before?

TheGreatDuck

4:59 AM

yup

ah, so you saw it, eh?

Eric Stucky

well, in that case, it represents a tremendous clarification of the concept :P

glad you were able to come to a nice formalization

TheGreatDuck

only took me way too long to see it was in my face the whole time. :p

Eric Stucky

haha, such sentences are only true in hindsight ^.^

TheGreatDuck

mmhm

to be fair though, I am also a programmer so sometimes it's the interface that is more important the implementation

Eric Stucky

I think most mathematicians would agree :P

I think that the conjecture you've posed is true, at least for reasonable DEs

I am toying around with a proof right now

TheGreatDuck

5:02 AM

actually, I think it is true for all the ODEs. When there are no continuous solutions, there are no solutions

Eric Stucky

But the principle is that, functions which have all their one-sided limits tend to be fairly reasonable.

TheGreatDuck

at least... on the interval.

well the unreasonable ones probably don't have solutions to begin with or are so obscure that they'd be an edge case to resolve.

hopefully it is the former and not the latter

Eric Stucky

oh, sorry, I misunderstood your last comment

yeah, if the domain is connected, I suspect it's true for all ODEs. That's reasonable enough :P

anyway, how is life otherwise?

school winding down for the semester, or do you have a while to go?

TheGreatDuck

finals start next week

and then that's it

Eric Stucky

sounds good :)

TheGreatDuck

5:10 AM

indeed it does

@EricStucky anything interesting lately?

Eric Stucky

Hmm, for me, yes.

I have narrowed my list of advisors down from 4 to 2, and I am confirmed reading with one of them this summer.

(commutative algebra)

So I am very optimistic about my progress in school :P

But mathematically, not particularly

somehow I ended up in four different classes which assigned no homework

I didn't respond to the free time particularly well :\

but I learned some tricks which I think will be nice when I get into thesis work proper jeje

TheGreatDuck

are you a grad student?

Eric Stucky

yes

TheGreatDuck

for some reason I thought you were a full professor. :p

Eric Stucky

haha, well according to my spam folder, you are not the only one under this impression -.-

TheGreatDuck

5:20 AM

your profile gives that impression

Eric Stucky

hmm, maybe I should change some of the verbiage then... I don't want to mislead people :P

So actually, I think I have a short answer to your question, at least for closed intervals.

TheGreatDuck

:-)

good answer or fatal flaw?

Eric Stucky

good answer :P

TheGreatDuck

yay

hey, at least it's better than one of my older beliefs.

for some reason, I thought at one point x*floor(1/x) was the integral of floor(1/x)

and that isn't even correct in the implied sense. :p

Eric Stucky

heh

TheGreatDuck

5:28 AM

in fact, I think it has no closed form but that's never been confirmed

based on 1 mod x having no closed form integral

actually, that would make sense given they differ by a constant and negation. XD

1 mod x = 1 - x*floor(1/x)

Eric Stucky

My answer is posted

TheGreatDuck

alrighty

:-)

Eric Stucky

It is perhaps not the most satisfying explanation, because it uses the big hammer of Darboux's theorem. But it also gives a fairly strong affirmative: D(y)=I(y).

TheGreatDuck

alrighty, but does that show there aren't any extraneous continuous solutions?

i.e. solutions to the I(y) that are continuous but not solutions to D(y)

(apologies if I'm being a bit dense here)

Eric Stucky

No, the question makes sense

TheGreatDuck

5:34 AM

ok

wasn't sure if the equality dealt with that

Eric Stucky

Yes, it does.

It shows that any continuous solution to I(y) is differentiable

TheGreatDuck

oh and does that extend to all differential equations?

Eric Stucky

Assuming the domain of a closed interval, yes

TheGreatDuck

sweet

and by closed interval you mean like [4,5] as an example, right?

Eric Stucky

right

so what I was getting at in the last paragraph is, obviously you want things like [4,5)

TheGreatDuck

5:36 AM

actually, I would like things like (-inf,inf)

Eric Stucky

oh...

TheGreatDuck

assuming existence of a total solution of course

Eric Stucky

I think that comes for free: something something locally compact

TheGreatDuck

i mean, if there isn't a solution on all real numbers then it should fail

Eric Stucky

right

TheGreatDuck

5:38 AM

i mean that simply because the differential equations class never restricted itself to domains and so preferably the ultimate solution would show that it works for all ODE solutions... whatever the bounds

but hey, it's a great start. Better than I could ever do. :-)

Eric Stucky

nah, certainly not, you just didn't know the magic word :P

TheGreatDuck

which is?

Eric Stucky

Darboux

TheGreatDuck

Definitely starting to seem like it was a lot easier to solve than I was thinking (but to be fair, I don't know analysis).

ah

i think the comments suggested such a thing but I didn't know enough to find it

:-)

at least now I can claim solutions to things without doubting myself. Of course, then one has to prove solutions to implied differential equations... but I'll let that slide as previous knowledge.

Eric Stucky

:P

I think those were pointing in rather different. In particular, I don't think you can extend my argument to say anything about subdifferential equations. Or perhaps it can be done, but it would take at least one extra step that doesn't spring to my mind.

TheGreatDuck

5:42 AM

i mean, I cannot prove the method of auxilliary equations gives all solutions anyways so why bother at the moment. I have a method now that does step function stuff. It's a decent optimization. Probably done before given the subderivative thing linked but I wasn't going for that anyways. I just want to solve diff eqs

Eric Stucky

Fair enough.

TheGreatDuck

@EricStucky Subdifferential was just a route to proving it. It might be a dead end like all the riemann hypothesis attempts. :p

Eric Stucky

jeje

TheGreatDuck

i mean, if the subderivative has extraneous solutions, then it's useless anyways.

and I dont know enough to say one way or the other

Eric Stucky

Oh, sorry, I still think the statement is true for SDE

TheGreatDuck

5:43 AM

I might play around with making the implied integral a more set in stone equation though

Eric Stucky

that would be nice :)

TheGreatDuck

thinking of maybe making it take in an (arbitrary) indicator function along with the function to be differentiated and that's how it chooses the side to use at each point

then it won't be multivalued

it'll just be a multivariate operator

i actually asked about the notation (not the statement as that would've been a bit irrelevant) to my abstract math professor and he suggested using ordered pairs instead of solution sets or systems of equations. I chose the 4th option. Just add another parameter. Makes everything work better when you do that. :p

Eric Stucky

Hmm, that seems reasonable.

(and that, also)

TheGreatDuck

of course, now the logical thing to do would be to examine fractional derivatives, PDEs (which are the one chapter we never touched in diff eq), and other stuff to see where else this might be extended into.

:-)

Eric Stucky

indeed ;)

TheGreatDuck

5:47 AM

not that it is incredibly useful, but it is fun.

at this point, I can definitely say that the one question that I've been editing over and over is now confirmed though

at least in some aspects

Eric Stucky

agreed =)

oh, okay!

I'll have to take a look at it again

TheGreatDuck

your answer just became the equivalent of beating it with a hammer

the other part to resolve is the finding of "continuous solutions", but I'm sure that falls under basic analysis of piecewise continuity.

I'm not going to accept yet though as it might auto award simply arts bounty. Don't want to do that to the guy.

in case he wants a better solution (one dealing with the edge cases and unclosed bounds)

no offense. :-)

Eric Stucky

my back is acting up again, so I'm probably off for the night.

none taken

see you around :)

TheGreatDuck

see ya

Axoren

Anyone familiar enough with exponential growth to answer this question? math.stackexchange.com/questions/2258664/…

Axoren

6:16 AM

@BalarkaSen You're up late.

What's up?

Balarka Sen

Up late? It's almost 12 in the morning :P

Not much.

Zee

7:28 AM

Is there anybody out there?

Alessandro Codenotti

Nobody home

Balarka Sen

i'm right here

Martin Sleziak

7:43 AM

I think of all the friends I've known. But when I dial the telephone. Nobody's home. (All by Myself)

Alessandro Codenotti

Mine was a Pink Floyd reference, "Is there anybody out there?" and "Nobody home" are the 2nd and 3rd track on the second CD of the wall

Balarka Sen

i have the wall on the list of things i should listen

Alessandro Codenotti

@Martin is every countable ordinal order isomorphic to a subset of $\mathcal{P}(\Bbb N)$ partially ordered by inclusion?

I think the wall is a masterpiece, but I also consider Pink Floyd the best group ever so I'm biased :P (I also went to Waters concert a few years ago and heard it live)

Balarka Sen

We're all biased here

lapalap

Hi!
Can anybody help me with identification of distribution of random vector?
its PDF is $p(x, y) = c e^{−(x+y)^{2}}$, where c is some constant

Alessandro Codenotti

7:59 AM

Integrate everything and adjust the constant so that the integral is 1

lapalap

@AlessandroCodenotti i need only type of distribution

Alessandro Codenotti

No idea then, I don't recognize it

Eric Stucky

8:17 AM

@lapalap: isn't that a Gaussian?

Astyx

Hi chat

Eric Stucky

hey ast

Lapa: Can confirm. It is a multivariate normal distribution on two independent variables (en.wikipedia.org/wiki/Multivariate_normal_distribution)

lapalap

@EricStucky isn't gaussian 2-d have $p(x, y) = c e^{-x^{2}/2 -y^{2}/2}$ ?

Eric Stucky

oh, yes. I did the freshman's dream thing :/

carry on

Alessandro Codenotti

It doesn't look like a vector of $2$ independent variables

keej

8:39 AM

Make a change of basis. rotate 45 degrees

Martin Sleziak

10:21 AM

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

Waiting

12:10 PM

@Hippalectryon how are you doing? :-)

Sha Vuklia

Guys, the unit sphere in $\mathbb R^n$ is defined as
$$
S^{n-1}=\{x\in\mathbb R^n:\Vert x\Vert=1\}.
$$
Why is it $S^{n-1}$ and not $S^n$? I'm guessing it has something to do with dimensions. When we look at the unit circle, $S^1$, it's clear that $S^1$ contains lines (which have dimension 1) - however, I can't quite see it in the generally case.

also, hi @Waiting!

Alessandro Codenotti

$S^n$ is an $n$-dimensional manifold

Astyx

In dimension 3, a sphere is a surface, ie an "object of dimension 2" and so on.

Alessandro Codenotti

$S^2$ is a surface ($2$ dimensional manifold), embedded in $3$ dimensional euclidean space

Sha Vuklia

so I would first need to understand what manifolds are before I can answer the question?

because then I will postpone it

Waiting

12:21 PM

@ShaVuklia Hey, great user @ShaVuklia! How are you doing? Math on Sunday? :D

I feel like I wanna watch a movie ... :P

Sha Vuklia

I mean, I can see it for $S^2$ and $S^3$, but I wouldn't really know what definition to apply for $S^n$. But I'll guess I'll wait until I know what manifolds are

@Waiting haha, yea I have too! I'm behind a lot :P

oh cool, I personally don't watch movies, but if you feel like it, have fun I guess :P

Astyx

You agree that a sphere is a surface and a circle a line ?

Sha Vuklia

yes @Astyx

Astyx

Just generalise that concept to higher dimensions

Sha Vuklia

hm yea ok. if the dimension of $\mathbb R^n$ goes up with one, then the dimension of $S^{n-1}$ will go up with one too?

Astyx

12:23 PM

It's like taking away one degree of liberty (the norm) away from your coordinate system in dimension $n$, resulting in something of dimension $n-1$

Of course that's not very rigorous explanation, but that's the idea

Sha Vuklia

ah okay, well that kind of makes sense, about taking away one degree of liberty!

Alessandro Codenotti

The intuitive idea is that every point of $S^2$ has a small circle around it that looks like $\Bbb R^2$

Waiting

@ShaVuklia I (pretty) rarely watch movies, but it happens sometimes. Last time I watched The Man Who Knew Infinity, a biographical film based on the book by Robert Kanigel (about the life of the famous Srinivasa Ramanujan).

Sha Vuklia

it's can't be dimension $n$ of course, otherwise it would be entire $\mathbb R^n$ :P

@Waiting oh, isn't that a new movie?

Waiting

@ShaVuklia Yeah, it is.

Sha Vuklia

12:25 PM

because I almost wanted to see that movie (when I was "pushed" to go to the cinema :P)

Alessandro Codenotti

@ShaVuklia uh, no, the open unit cube is a 3 dimensional submanifold of $\Bbb R^3$

Astyx

For instance in spherical coordinates in dimension $3$, you have $r, \theta, \phi$ an to get the sphere, you take $r=1$ so you only have two remaining coordinates

Semiclassical

hi chat

Sha Vuklia

@AlessandroCodenotti oh okay, well I stick to what I know now then for $n=2$ and $n=3$ :P without considering (sub)manifolds :d

hi @semi

Semiclassical

another way to say it is that if you look at some portion of a globe, you can treat the lines of latitude and longitude as 'xy' coordinates

(this isn't actually different than the spherical coordinates remark, but it's rather concrete)

Waiting

12:29 PM

@ShaVuklia Really? :P I like it very much, there are some memorable parts in the movies, especially the discussion between Ramanujan and Hardy about the way of viewing mathematics, that is like an art.

Ramanujan was first an artist, has brought mathematics to the art level, it wasn't just about solving problems.

Balarka Sen

@Sha In $\Bbb R^n$ you have $n$ degrees of freedom. When you look at the set $x_1^2 + x_2^2 + \cdots + x_n^2= 1$ (namely, $S^{n-1} \subset \Bbb R^n$), that is like having $n$ degrees of freedom but there is a 1 dimensional restriction (because it's a single equation the set is cut out by). So there are $n - 1$ degrees of freedom.

I am not sure how helpful that is but it's a way of thinking.

In general the picture is exactly like $S^2$ in $\Bbb R^3$.

Semiclassical

It goes to the same counting you have in linear algebra: n unknowns - k constraints = n-k dimensional space of solutions. (setting aside linear dependence issues etc.)

That linkage can be made precise, i suppose, by linearizing the equation $\sum_{k=1}^n x_k^2=1$ about some specific point.

Lozansky

If I have a coordinate system $\{u,v,w\}$ where $u,v,w$ are functions of a spherical coordinate system $\{r, \phi, \theta\}$, how do I find the scalars $h_u, h_v, h_w$?

Balarka Sen

@Semiclassical Right (what I am saying is just the preimage theorem but I don't want to say it out loud)

Fargle

To play devil's advocate, while I wax poetic about the beauty in mathematics as much as the other guy, I think to call it either an art or a science ignores its true practice and function.

Semiclassical

12:34 PM

Right. @BalarkaSen

Lozansky

I know that $h_i = |\frac{\partial \mathbf{r}}{\partial u_i}|$

Sha Vuklia

@Balarka thanks! that's helpful :)

@Waiting and yea, I’m sure that conversation was nice, but to tell you the truth, the reason I didn’t feel like watching it, was because the movie contained all these male academics (and one Indian girl, who was his gf?) - and I thought by myself, why would I watch it, when I can’t relate anyways? It’s not my world:d

Lozansky

But that doesn't seem to lead anywhere since I don't know a non-tedious way to express the position vector $\mathbf{r}$ in terms of $u,v,w$

Semiclassical

You might find the movie 'Hidden Figures' interesting.

(full disclosure, I haven't seen it myself.)

Waiting

@ShaVuklia loll, that was a cute observation. :-) Still, the life of Ramanujan is entirely fascinating, letting apart how the movie is presented, it's perhaps the most romantic mathematical story in the history (it was said that already).

Lozansky

12:38 PM

Heya, @Semiclassical

Semiclassical

hi

Balarka Sen

I watched the 1982's The Thing a few days ago

cool stuff

Semiclassical

@Lozansky Can't you use the multivariable chain rule here?

Waiting

@ShaVuklia I have to go now. Take care. :P

Sha Vuklia

@Waiting alright, take care!

Lozansky

12:40 PM

@Semiclassical Not sure how though... let me post the coordinate system in question

Semiclassical

e.g. $\displaystyle \frac{dx}{d\phi}=\frac{dx}{dx}\frac{\partial x}{\partial \phi}+\frac{dx}{dy}\frac{\partial y}{\partial \phi}+\frac{dx}{dz}\frac{\partial z}{\partial \phi}$

Lozansky

$\cases{u=r\sin^2 \frac{\theta}{2}\\ v=r\cos^2\frac{\theta}{2}\\w=2\phi}$

Semiclassical

Hmm.

Lozansky

Wait... isn't $h_i = \frac{1}{| \nabla u_i|}$?

Semiclassical

tbh I don't remember

Lozansky

12:45 PM

It makes sense since $\nabla \phi = \sum \frac{1}{h_i} \frac{\partial \phi}{\partial u_i} \mathbf{e}_i$

I think that's correct

Semiclassical

right.

Lozansky

It gives me the correct answer so I'm assuming it's right :P

jserv

Hey all. I'm taking an economics course, one of the equations representing a production function has a 'min' in it. Specicially it looks like h(m,c) = (min{m,c})^1/2. I'm really not sure what this 'min' does in this context. Can anyone help me?

Semiclassical

outputs the smaller of m,c

jserv

So if m = 40, and c = 10, it will always be h(m,c) = 10^1/2?

Semiclassical

1:02 PM

Should be.

But if you decrease m below c, then it'll be m^(1/2) instead.

jserv

m and c are defined from the outset, so this really shouldn't change regardless of how many units are produced, right?

Semiclassical

right.

jserv

thanks, @Semiclassical

Secret

[Super Random] Holy grail of circumvention:
$$\int_{\textrm{all}}\textrm{Savage }d[\mu]=0$$

Therefore:
$$\lim_{\textrm{Perfect}}\textrm{Circumvention}=\textrm{Circumvention}^{-1}$$

Or in words: The most perfect circumvention given a scenario is blocking all possible means to circumvent it

jserv

1:29 PM

I feel like I must have misread this question. The production function as I'm using it just continues to put out an ever greater profit with however many units you make, but the question asks you to solve for an optimal number of units. This is the entire question: i.snag.gy/qEG76h.jpg

Tim The Enchanter

1:42 PM

@jserv Perhaps it means marginal profit, or the profit to cost ratio. en.wikipedia.org/wiki/…

Lozansky

1:59 PM

@jserv Maybe solve $0.4m + 0.1c = 4min(m,c)^.5$

Secret

2:19 PM

That post (and a few earlier ones above that due to at that time there's an active conversation) showed how given any continuous function that is a solution to an ODE must be a solution to both the left and right ODE (implied ODE in your terminology)
, but the converse is not necessary true, due to the complication that you can have continuous functions that obey the uniform convergence criteria outlined in the MSE post in my message, with discontinuous one sided derivatives even when derivatives exists everywhere

RaGe MaGiXZ

Anybody know ow to work out the area under a 3 dimensional curve shape

Secret

55

Q: Discontinuous derivative.

Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.

real-analysis derivatives

I am not sure whether these functions actually obey the uniform convergence criteria in this MSE link:

yesterday, by Secret

7

Q: Under what conditions can I interchange the order of limits for a function of two variable?

Suppose I have $f:\mathbb{R}^2 \to \mathbb{R}$. What conditions do I need to say that $$\lim_{x \to a} \lim_{y \to b} f(x,y) = \lim_{y \to b} \lim_{x \to a} f(x,y)$$ ? What about in a more general case, by taking $X,Y$ and $Z$ topological (Hausdorff) spaces and $f$ from $X \times Y$ to $Z$ ?...

real-analysis general-topology limits

but the point is, if your solution to either your left or right ODE are everywhere continuous and everywhere differentiable functions given in the above link (and I suspect something like $x^2 \sin (\frac{1}{x})$ may be a solution that can arise in some oscillatory physical systems), then you must check whether it also satisfy the other one-sided ODE (i.e. right or left) otherwise it won't be a solution to the original (two sided) ODE

That is, even if $f$ is continuous and differentiable, the following
$\textrm{f solves either $D_+(y)=c$ or $D_-(y)=c$}\implies \textrm{f solves $D(y)=c$ }$
does not hold in general

Semiclassical

2:36 PM

@jserv The way I read that is that, if you have 9 cups of milk and 15 tablespoons of cocoa powder, you'll only be able to use 9 of those tablespoons of cocoa to produce 3 cups of hot chocolate.

jserv

@Semiclassical That's the entire question, no limit on the number of ingredients is mentioned

Semiclassical

It's implied. If I don't have enough milk, all the cocoa powder in the world won't help me produce more hot chocolate.

Secret

Hmm, what is the laplace transform of $x^n \sin {\frac{1}{x}}$...?, if the outcome has no step functions, then the implied ODE formalism will work because such example will never arise in the implied ODE formalilsm, and thus solution there will automatically guarentee solution to the original ODE

gets wolfram alpha

Semiclassical

Working in units of dollars, the profit you make is given by $p(m,c)=4h(m,c)-0.4m-0.1c$. @jserv

Balarka Sen

Hi @PaulPlummer

Paul Plummer

2:42 PM

@BalarkaSen Hello

Semiclassical

If I assume that I've got more milk than cocoa, then my production is determined by the cocoa as $h=\sqrt{c}$. In that case if $c$ is large enough the cost of buying cocoa will exceed the profit I actually make from it. (The cost grows as $c$ and the revenue as $\sqrt{c}$.)

Same goes if I have more milk than cocoa.

So there should be an optimal amount of cocoa and milk that I can buy to get the best profit.

Hippalectryon

@Waiting (sorry, I was away) fine, and you ?

jserv

If you had more cocoa than milk, would the production change from h = sqrt(m)? Would that give a different value?

Change to*, rather

Semiclassical

Right.

However, do you want to have more cocoa than milk? (or vice versa)

Paul Plummer

What have you been up to @BalarkaSen

Semiclassical

2:50 PM

It's not obvious that you do, since (using the analogy I gave) some of the cocoa will end up going to waste if you have too little milk.

Astyx

Hi @Hippa !

Hippalectryon

@Astyx \o

Semiclassical

I think that may depend on the relative costs, though.

Astyx

Alors la reprise ?

Hippalectryon

@Astyx Ca va, c'est assez tranquille pour l'instant

Astyx

2:51 PM

Vous avez des amphis c'est ça ?

Hippalectryon

Des amphis et des petites classes

jserv

Thanks @Semiclassical , I'm struggling to know how to apply this, though. If I take your equation from earlier, right a bunch of possible different values for h, would I eventually get the point where you get the greatest p? Or am I going about this the wrong way?

Balarka Sen

@PaulPlummer I should work on foliations but I am procrastinating.

Semiclassical

Well, the simplest way I know to get a handle on what's going on is to make a contour plot.

To that end:

The horizontal and vertical axes are the units of milk and cocoa bought respectively, and the contours are level sets of the profit function I gave earlier.

The blue is profit less than zero. The other contours go up in steps of 1.

Based on that, the optimal result should be somewhere around 17 units of both cocoa and milk.

If you can argue that you indeed need $m=c$ for the optimal profit, then you just have a one-variable optimization problem. But it's not immediately obvious to me why that should be true.

Transcript for

$Mathematics$ Mathematics