Mathematics - 2017-11-08 (page 1 of 3)

r9m

00:40

Can a holomorphic function on the open unit disc have an annulus as it's image?

MATH ASKER

Sand being dumped from a funnel forms a conical pile whose height is always one third the diameter of the base. If the sand is dumped at the rate of two cubic meters per minute, how fast is the pile rising when its 1 meter deep?

So i have the volume of the cone which is $V=1/3πr^2h$ and $dv/dt = 2 meters cube$ is given I'm having trouble after this

should i take the derivative of the volume formula?

Secret

01:11

that's exactly what you need to do, since V is a function of h, thus dV/dt is contributed by dh/dt

Secret

01:40

Materials that exploit topology in some way are interesting, because they fit my preference for something that apparently not having some property A but it actually does

nature.com/articles/nmat4998

For example, a porous liquid is something that you thought it is very crowded, hence not much space, yet there is plenty of space to hold molecules

It's almost like the concept of a TARDIS, being bigger on the inside

More generally, I found concepts that packs a lot of things in a small space (or even no actual spatial dimensions at all) interesting

James

Hello good evening, is @TedShifrin back?

Secret

This is why Skyrmions are interesting, because you have so much information packed in such a small space

Simply Beautiful Art

An intro to ordinals and how they can be translated into large finite numbers.

Secret

01:55

in Simply Beautiful Art's realm of calculus and analysis, 12 hours ago, by Simply Beautiful Art

@Mr.Xcoder ω_n is the nth ordinal with no bijection to lower ordinals... I can explain that later, but its not important here.

This one is no fun because we cannot even write it down explicitly even using countably many turing jumps

Simply Beautiful Art

lol

I mean

countably (pretty much) anything only takes you up to ω_1.

Secret

but not $\omega_1$ itself

in Mathworks (Not the main chat!), 21 hours ago, by Secret

Right, what about countable alphabet with countable string length (but such string can only be or order type $\omega$, with each letter possibly interspaced by only operators?

and we cannot even do that even if we have a countable alphabet with countably long strings

in Mathworks (Not the main chat!), 21 hours ago, by user21820

Because there are uncountably many strings of length ω.

in Mathworks (Not the main chat!), 21 hours ago, by user21820

And practically all of them are indescribable.

in Mathworks (Not the main chat!), 21 hours ago, by user21820

My point is: It's useless to say "notation" if it is not computable in some sense.

Simply Beautiful Art

:P

Faust

Mornign

Simply Beautiful Art

Looks like a start of a long morning for you.

Faust

02:01

i got a painful geometry assinment to finish

over like the next 38hrs or so

Simply Beautiful Art

And I've got persuasive speech outlines due tomorrow.

Faust

gl

Simply Beautiful Art

yeep.

Secret

Currently doing some coding in the background to extract my raw data for analysis in my chemistry

googology.wikia.com/wiki/Really_Big_Ass_Number

Lol

Simply Beautiful Art

x'D

Attempted to make an uncomputable function?

Wow, big dreams there.

Secret

02:06

Cannot make uncomputable function without solid understanding of computability, which I am still poor

Uncomputable functions Edit
These functions are uncomputable, and cannot be evaluated by computer programs in finite time.

Busy beaver function
Frantic frog function
Placid platypus function (slow-growing)
Weary wombat function (slow-growing)
mth order busy beaver function
Betti number
Doodle function
Xi function
Infinite time Turing machine busy beaver
Rayo's function
FOOT function

Simply Beautiful Art

@Secret Means what it says. A computer can't compute it. :P

Secret

Besides, since in my "world" countably infinite strings with order type $\omega$ exists, to me uncomputability is actually 2nd uncomputability

i.e. cannot give result in countable time

Simply Beautiful Art

welp

I shall head to bed.

Good night all

Secret

> It is easy to simulate a TM, but it is much harder to predict the behavior of a TM. This is because some TMs never halt, and the only way to test for that is to A) simulate them infinitely, or B) find and prove a pattern. Option A is physically impossible, and option B is difficult — how does a computer recognize arbitrary patterns? — and also problematic because many TMs exhibit chaotic behavior and do not have simple patterns.

googology.wikia.com/wiki/Infinite_time_Turing_machine

10000... = 0

01000... = 1

00100... = 2

00010... = 3

...

Constructing $\omega$

00000000000000000000000000...

01000000000000000000000000...

01100000000000000000000000...

01100100000000000000000000...

error

00000000000000000000000000...

01000000000000000000000000...

01000010000000000000000000...

01000010010000000000000000...

error

00000000000000000000000000...

01000000000000000000000000...

01001000000000000000000000...

01001001000000000000000000...

01001001001000000000000000...

...

01001001001001001001001001... = $\omega$

Constructing $\omega + n$

00100100100100100100100100...

00010010010010010010010010...

00001001001001001001001001...

00000100100100100100100100...

00000010010010010010010010...

00000001001001001001001001...

...

Constructing $\omega 2$

00000000000000000000000000...

Jenna Sloan

02:42

Error: MessageOverflowException

Secret

System: Transferring unsaved progress to New Room

user472288

03:05

Is this new?

Jenna Sloan

@user472288 Is what new?

user472288

Chat Rooms in SE

Jenna Sloan

No,

user472288

Oh

Must have missed it.

Any mathematicians around?

Wildcard

03:44

@user472288 Evidently not.

Balarka Sen

08:46

Hi chat

Tobias Kildetoft

Hi

user21312

How can one show that p(b)=p(b intersect a complement)+p(b intersect a)?

Balarka Sen

09:13

This is a million times better than the winner

If there is any good in the world, this needs to be on the star panel

I can't stop listening to it. This is too good.

I unironically love it

Alessandro Codenotti

09:29

How does one prove that if $G$ is a finite group and $p$ is a prime dividing its order then there is an element of order $p$ in $G$?

Balarka Sen

This is Cauchy's theorem, I believe

Tobias Kildetoft

@AlessandroCodenotti As Balarka said, this is Cauchy's theorem. There are about a million different proofs

None of them simple enough that I would expect a student to work it out without many hints

Balarka Sen

Look at the center Z(G). If p divides |Z(G)|, then Z(G) contains a Z/p factor because it's abelian, and this is easy to prove for abelian dudes

So look at G/Z(G); p has to divide the order of that

Tobias Kildetoft

@BalarkaSen So what happens when the center is trivial?

Balarka Sen

Ya that's why I quotiented out by the center

Tobias Kildetoft

09:32

that doesn't change anything if it is trivial

Balarka Sen

Hmm, I'm not sure I remember how this goes

@Tobias No, I mean, the problem boils down to proving it for zero center groups

Oh hm I guess I use the class equation

Tobias Kildetoft

@BalarkaSen Right. That will do it, when done correctly (by induction)

Balarka Sen

|G| = |Z(G)| + (sum of order of the various conjugacy classes)

Hm

Alessandro Codenotti

@TobiasKildetoft I see, I guess I'll look it up then

Actually I have a group where every element has order a power of a prime $p$, is there an easy way to show that it is a $p$-group without invoking Cauchy's theorem?

Tobias Kildetoft

@BalarkaSen Consider a minimal $G$ which has order divisible by $p$, but no element of order $p$

Alessandro Codenotti

09:36

(Everything is abelian if that helps)

Tobias Kildetoft

@AlessandroCodenotti once it is abelian, Cauchy's theorem is easy by induction. If it is cyclic it is trivial, and otherwise, take an element and quotient by the subgroup it generates, then do induction

user84215

in Abstract Algebra Course, Mathematics and Physics University, Oct 25 at 10:03, by MathematicsAminPhysics

Sketch of proof of 2.11.: The proof proceeds by induction on $|G|$. Suppose $x$ is an element of $G$ such that $p$ does not divide $|x|$. Let $N= \langle x \rangle$. By Lagrange's theorem, $|G/N|<|G|$. Since $p$ does not divide $|N|$, we must have $p \, \big {|} \, |G/N|$. By applying the induction assumption to the smaller group $G/N$ one can conclude that it contains an element, $\bar {y} = yN$, of order $p$. Also we have $\langle y^p \rangle \neq \langle y \rangle$, that is, $|y^p|<|y|$.

Balarka Sen

@Alessandro If the group is abelian, just nuke it by classification of finite abgroups

(It can be done without, but why not)

Tobias Kildetoft

@BalarkaSen Otherwise, nuke it with Sylow

Balarka Sen

Hah

Alessandro Codenotti

09:38

@BalarkaSen we need it as a preliminary step for that classification :P

Balarka Sen

oh well rip

It's not hard to prove it in any case

@TobiasKildetoft Hmm, OK. So G = 1 + (sum of orders of various conjugacy classes) implies one of the conjugacy classes must not have size divisible by p, right?

Tobias Kildetoft

@BalarkaSen Right

Balarka Sen

I guess that means by orbit-stabilizer theorem the stabilizer of that representative is going to be a subgroup of G of order divisible by p

And that breaks minimality

Tobias Kildetoft

right

Leaky Nun

10:13

78 79 84 73 67 69 32 78 79 84 73 67 69

(I’m looking for trouble...)

@TobiasKildetoft but that isn’t constructive is it

Balarka Sen

It's an induction argument, where you induct on the order of the group.

I also don't know what constructive means in this context. That's for statements of the form "Such an object with such and such properties exist/does not exist"

Leaky Nun

but it uses contradiction

Balarka Sen

@LeakyNun It doesn't need to use contradiction.

Leaky Nun

it starts with “if it is false”

Tobias Kildetoft

@LeakyNun But why would we care if it is constructive?

Leaky Nun

10:20

and is continued with “then we can find a minimal example”

Balarka Sen

I leave it as an exercise to you to not use that. It's a simple modification.

Like I said, it's an induction argument.

Leaky Nun

you still need contradiction in the inductive step

Balarka Sen

Assume for all groups of order $\leq n$, $p$ divides the order $\implies$ there is a subgroup of order $p$. Pick a group $G$ of order $n + 1$, and go through the argument to find a subgroup of $G$ (an appropriate centralizer) whose order is divisible by $p$.

Leaky Nun

@TobiasKildetoft because I’m pointless

Balarka Sen

But that is a proper subgroup of $G$, hence of order $\leq n$.

By induction that has a subgroup of order $p$

Hence this statement is true: "For all groups of order $\leq n +1$, $p$ divides order $\implies$ there is a subgroup of order $p$"

Induction finishes it.

Where do you need contradiction for this?

Leaky Nun

10:24

nice

Balarka Sen

Yes, learn to think before commenting.

James

10:45

Hello again...if $\{X, X', X \times X'\}$ is an orthonormal basis, is $X \cdot X''$ zero?

Tobias Kildetoft

@James There is no $X''$ there

James

Does it follow that $X$ and $X''$ are orthogonal, or parallel, or we can't conclude?

Tobias Kildetoft

there is still no $X''$

James

@TobiasKildetoft

user84215

@James Hello.

James

10:47

Hello again, @MathematicsAminPhysics

I am still not done with the problem, but I have leapt strides

Tobias Kildetoft

@James The setup has no mention of $X''$, so how could you know anything about it?

James

So in short, we can't conclude anything about $X \cdot X''$?

Tobias Kildetoft

@James No, how could we?

James

How about $X' \cdot X''$? Is this zero?

Tobias Kildetoft

@James How is that question different from the last one?

James

10:50

Could $\{X', X'', X' \times X'' \}$ be another orthonormal basis?

Tobias Kildetoft

@James Yes, of course, since $X''$ could be anything for all I know

James

@TobiasKildetoft, I'm actually talking about this problem, I'm sorry if I didn't give a proper introduction

0

Q: Gaussian and Mean Curvatures for a Ruled Surface

We are asked to prove the following theorem found in page 88 of Differential Geometry: Curves, Surfaces, Manifolds by Wolfgang Kühnel. Using standard parameters, calculate the Gaussian curvature and the mean curvature of a ruled surface as follows: $K = -\dfrac {{\lambda}^2} {{{\lambda}^2 +v^2}...

Tobias Kildetoft

@James Sorry, but I don't see how anyone could have inferred that amount of background from this

James

Anyway, let me put down my solutions
$c' = FX + \lambda(X \times X')$ so $c'' = F'X + FX' + \lambda ' (X \times X') + \lambda (X \times X')'$

Simplifying $c'' = F'X + FX' + \lambda ' (X \times X') + \lambda (X \times X'')$

so $C'' \cdot X = F'X \cdot X + FX' \cdot X + \lambda ' (X \times X') \cdot X + \lambda (X \times X'') \cdot X' = F' + \lambda ' (X \times X'') \cdot X = F'$

Leaky Nun

@Astyx is my observation correct that “situé” is pronounced differently than “si tu es” because the “t” in the former is more palatalized?

James

10:58

Similarly,
$c'' \cdot X' = F - \lambda (X'' \cdot X \times X')$
$c'' \cdot (X \times X') = \lambda + \lambda (X' \times X'')$

@TedShifrin, am I on the right track?

Balarka Sen

11:18

@Tobias I think these are all derivatives.

In which case, $X$ and $X'$ being orthogonal means $X \cdot X' = 0$, taking derivative of which, we get $X' \cdot X' + X \cdot X'' = 0$, hence $X \cdot X'' = -1$ as everything's a unit vector in orthonormal bases.

Hi @TastyRomeo

user84215

12:04

The following workshop will start in the Mathematics Workshops room:

user84215

Finite Group Theory, at 9:30 GMT on Sunday, November 12, 2017.

user84215

in Mathematics Workshops, 3 mins ago, by MathematicsAminPhysics

In this workshop we want to study structure of many examples of finite groups by using basic theorems in finite group theory like Lagrange's theorem, Sylow's theorem, ... and basic tools like direct product, semidirect product, ... (You can find them in Abstract Algebra Course, Mathematics and Physics University).

Secret

I am ok with you posting announcement to you room, but please get rid of NOTICE NOTICE so that people will not be mistaken it is an MSE official anouncement

Journeyman Geek

Hi - any local mods hang out here?

Secret

thank you

Balarka Sen

12:07

Not many that I know of

Journeyman Geek

ahh ok

Secret

The algebra course does has its use, though I am less optimistic about the general topology course one

Journeyman Geek

Is this chap posting notice in all caps and posting links to his room a problem?

Balarka Sen

mixedmath has been regularly patrolling recently, and ACuriousMind comes around once in a while

@JourneymanGeek He has been flag-banned before for doing that

Here

Journeyman Geek

(chat mods are mods everywhere, however, I'd like to spend a little time knowing if its an issue)

Balarka Sen

12:08

So I suppose it is generally perceived as a problem

Journeyman Geek

Ah, not quite what I asked, but close enough

Leaky Nun

The user has already been suspended for 6 days, so please keep our discussion minimal (or better, non-existent).

Secret

[Random]

"Prime number base number system"

Journeyman Geek

@LeakyNun oh that was me

and I've basically let a maths mod know.

Well, I guess that should deal with that, until someone more familiar with the chatroom can take a look. Cheers, and sorry for any distress caused ^^

Balarka Sen

See ya

Secret

12:23

Let the position bases of this system be the prime numbers in its usual increasing well order

{2,3,5,7,11,13,17,23,29,31,37,...}

Now, each position can be either 0 or 1, to indicate whether that prime is present or not

For example:

{1,0,0,0,0,0,0,0,...} = 2
{1,0,0,1,0,0,0,0,...} = 2*11=22

Now to complete the sentence, I need to dig the chat log for something...

Secret

12:44

hmm, wait a sec.., this might not work

Even if we did have basically two lexicographic systems nested in each other and all numerals are unique because of the fundemental theorem of arithmetic, the issue lies on the unbounded sequences: e.g.

{1,0,1,0,1,0,1,0,...}
{1,0,0,1,1,0,1,0,...}

there is simply no way to compute these unless we allow countable sequences

Btw, the two lexicograph ordering is as follows:

Lexicographic ordering one is binary such that given the sequence where the positions are arranged in increasing primes, then since each entry can be either 1 or 0,we basically have a binary representation of natural numbers

Lexicographic ordering two is the interesting one: For every binary sequence being written down, compute the required product of primes (which is bijective to the binary sequence owing to the fundamental theorem of arithmetic). The larger number is the one with larger final product

Combining these two orderings, given two binary sequences A and B, we should be able to determine whether A > B o A < B by using lexicographical ordering 1 > lexicographical ordering 2

To illustrate:

{1,0,0,0,0,0,0,0,0,...} < {0,1,0,0,0,0,0,0,0,...} < {1,1,0,0,0,0,0,0,0,...} < {1,0,1,0,0,0,0,0,0,...} by 1st lexicographical orderuing

actually wait a minute, since the positions are a monotonically increasing sequence of primes, it follows that the two lexicographic orderings are isomorphic to each other

user21820

@Secret: I did not read most of what you wrote earlier in here.

But it's very simple. You just have to work logically instead of guessing.

You want an order-preserving bijection f from Q to Q ⋃ {r} such that f(0) = r.

Secret

uh, the above block of text is a completely different thing, I have not figure out the rational mapping yet

user21820

Well so that means that f maps (−∞,0) to (−∞,r).

You should know that Q ⋂ (−∞,0) is order-isomorphic to Q ⋂ (0,1).

Secret

13:01

yup, and ln (x) can do that mapping. The problem is the mapping to r, I don't know of any function that does not map irrationals to rationals and vise versa

user21820

@Secret See my edited message.

ln is useless here.

And in general Q ⋂ (0,1) is order-isomorphic to Q ⋂ (p,q) for any rationals p,q such that p < q.

And the message just before my edited message should also be restricted to Q; I can't edit it anymore.

And Q ⋂ (−∞,r) splits into ( Q ⋂ (−∞,0) ) ⋃ ( Q ⋂ [0,r) ).

So you already can make the first part from Q ⋂ (0,1/2).

And then you need to make the second part from Q ⋂ [1/2,1).

So you want f(1/2) = ?

Then just ask yourself what you want for the following:

f(2/3) = ?

f(3/4) = ?

...

You just need to make sure you end up covering Q ⋂ [0,r).

Secret

hmm...

James

13:22

@TedShifrin, I am done! The longest was computing $\langle f_u \times f_v, f_{uu} \rangle$

Maxime

What kind of physics questions allowed?

A parallel beam of light is incident from air at an angle a on the side PQ of a right angled triangular prism of refractive index n=\sqrt {2}.
Light undergoes total internal reflection in the prism at the face PR when a has a minimum value of 45°. The angle θ of the prism is

I wanna ask this question on math.se

if that's mathematical physics

Akiva Weinberger

@Maxime Do you know Snell's law

TastyRomeo

13:38

If your question is purely about the mathematical calculations involved in a physics problem, it's probably fine for MSE

But if your question relates to a physical law or how to apply it, it's probably not and should be on Physics SE

Maxime

@AkivaWeinberger

Exactly, I know it but not too much.

@TastyRomeo Can you show some examples for it?

0

Q: What is the angle $\theta$ of the prism?

A parallel beam of light is incident from air at an angle $a$ on the side $PQ$ of a right angled triangular prism of refractive index $n = \sqrt {2}$. Light undergoes total internal reflection in the prism at the face $PR$ when $a$ has a minimum value of $45°$. What is the angle $\theta$ of the...

calculus physics triangle mathematical-physics

Also I've posted my question a second ago.

You can take a look.

Faust

14:13

Coconut

GFauxPas

Is there a link between AM-GM and the propety of eigenvalues that the arithmetic multiplicity $\ge$ geometric multiplicity? Or is it just a coincidence that their names are similar

anon

14:39

...you mean algebraic multiplicity?

no relation

GFauxPas

yeah algebraic I meant

ah, well, its a mneumonic at least

Secret

e.g. f(1/2)=0, f(2/3)=1/2, f(3/4)=2/3,... f(1/3)=2/5, f(2/5)=3/7,..., which at best is a piecemeal function with countably many cases of the form $f(p/q)=\frac{p+k}{q+m}$ where there is a g(p,q)=(k,m) with g cannot be written down in finite number of steps

hmm... if I need to end up specifying the image of f for each rational individually and there is no finite string e.g. f(x)=some finite strings of operations, then it will mean that f is uncomputable?

user21820

@Secret Look. Where in this did you ensure what is stated in the message you replied to?

The whole point is that you need to cover all rationals in [0,r), and you're not doing it. And no, what I said earlier stands; if you choose a reasonable irrational r the same construction will get you a computable map.

Secret

I know f(1/2)=0 and $\lim_{x\to 1}f(x) = \lim_{y\to r} y$ based on the way the interval is partitioned as described in the previous message. But so far I failed to find any continous function f(x)=some finite number of +,-,*,/, trancendental function applied on x that will not map at least one rational to the irrationals

user21820

Have you even chosen r?

Even if you don't, you can still choose reasonably nice values for f on the points I stated.

flow

14:51

Hi can anyone tell me if this is right: A subset of B and B subset of C -> A element of C I read this at math.stackexchange.com/questions/2504991/…

Secret

I tried $r={}^3\sqrt{2}$ and I tried polynomials, combinations of trigs but all of them failed

user21820

@Secret Why that instead of just sqrt(2)?

Akiva Weinberger

@Faust Is that all

Faust

?

Akiva Weinberger

Coconut?

user21820

14:52

@Secret: And that's not the point. Just pick in a systematic way the values of f that I asked you for.

Faust

it was 5am cut me some slack

Akiva Weinberger

There's two ways to do it, there's the "Just Do It" approach that @user21820 is suggesting

and there's a more explicit approach

user21820

@AkivaWeinberger Actually the obvious just-do-it approach should lead to an explicit map.

Secret

$f(\frac{1}{2}) = 0, f(\frac{2}{3}) = \frac{1}{2}, f(\frac{3}{4}) = \frac{2}{3}$

Balarka Sen

Shia Labeouf approach, you mean

user21820

14:54

@Secret This is not systematic; you didn't show how to continue.

Worse still, it won't work if r < 2/3.

@AkivaWeinberger: Explicit enough that if r is computable then the map would be computable.

Secret

$f(\frac{q}{q+1}) = \frac{q-1}{q}, q\in \Bbb{N}$ denominator nonzero

= FAIL

user21820

@Secret If you cannot reason abstractly about an arbitrary irrational r, pick a fixed one and then make f work for that r.

Akiva Weinberger

Maps backwards is spam

2

user21820

@AkivaWeinberger Lol!

Secret

Ok let's pick $r=\sqrt{2}$ again

Given $[1/2,1) \to [0,\sqrt{2})$ we knew the following:

$f(\frac{1}{2})=0, \lim_{x\to 1} f(x) = \lim_{y\to \sqrt{2}} y$

so now to find... $f(\frac{2}{3}), f(\frac{3}{4}),...$

user21820

15:01

You can't 'find', but you can systematically pick.

Secret

$\sqrt{2}\approx 1.41421356237...$

$f(\frac{1}{2}) = 0, f(\frac{2}{3}) = 1,f(\frac{3}{4}) = 1.4,f(\frac{4}{5}) = 1.41, f(\frac{5}{6}) = 1.414,...$

$f(\frac{1}{3}) = ?$

Thinking of an f that defines a countable sequence taking prime power positions of the decimal representation of $\sqrt{2}$ which grows slower as the denominator of the initial fraction $\frac{1}{n}$ increases...

Akiva Weinberger

15:18

Baire Spaces: Space Bears

https://i.imgur.com/vJM8lqB.jpg

Secret

and therefore, what to do now is to find a countable sequence of functions {$g_0,g_1,g_2,...,g_n...$} that grows slower as the index $n$ increases

Akiva Weinberger

lol

Secret

and $g$ must preserve rationals and not go beyond $\sqrt{2}$. Hmmm....

Maneesh Narayanan

$f$ is a polynomial of degree atmost 3. right?. Then how would be the $f(z)$ takes the form of a series.math.stackexchange.com/questions/403031/…. Please explain the answer given below by Ayman

Secret

Akiva Weinberger

15:25

@ManeeshNarayanan $a_n=0$ for $n>1$ then

Maneesh Narayanan

@AkivaWeinberger Thank you. Let me read again.

Akiva Weinberger

lol: https://i.redd.it/p61id6nmatuz.jpg

Faust

thats just wierd

Maneesh Narayanan

@AkivaWeinberger Let $f(z)=a_0+a_1z+a_2z^2+a_3z^3$ and given that $f(z)=f(iz)$ \implies $a_1=a_2=a_3=0\implies f $ is a constant. am I right?

Akiva Weinberger

Ya

Secret

15:34

This family is growing in the wrong direction

Maneesh Narayanan

@AkivaWeinberger Thank you.

Secret

Ok I need to sleep. There is no increasing sequence of functions that preserves rationals and with decreasing rate of growth

Blodstone

Hi, is it common to write a function of $f(x,y)$ as $f_{x,y}$

Liad

$f:\Omega \to \Bbb R , \Omega\subset \Bbb R \ ^ 2$. i need to prove that if $f$ has continuous partial derivatives then $F(x) = \int_a^b f(x,t)dt$ is differential and $F' = \int_a^b \dfrac{\partial f}{\partial x} dt$ . the hint state that i should use Lagrange intermediate value theorem and the fact that the partial derivative is continuous in order to show that for each $\epsilon \gt 0 $

there is $\delta \gt 0$ s.t $|\dfrac{f(x+h ,t) - f(x,t)}{h} -\dfrac{\partial f(x,t)}{\partial x} | \lt \epsilon$. im not sure why should i use Lagrange intermediate value theorem ? doesn't it follows immediately by the def. of the partial derivative and the fact that it exists ?

Secret

15:50

37

Q: Functions that take rationals to rationals

What is known about $\mathcal C^\infty$ functions $\mathbb R\to\mathbb R$ that always take rationals to rationals? Are they all quotients of polynomials? If not, are there any that are bounded yet don't tend to a limit for $x\to +\infty$? If there are, then can we also require them to be analytic...

real-analysis

Akiva Weinberger

$\cal C^\infty$?

Secret

[Unrelated random] Just before I forgot

Define positions with values {2,5,3,7,13,11,19,17,...}. Let ordering 1 be increasing from left to right
Define numerals 0,1 for each position. Let ordering 2 be increasing when given binary sequences b and c. b > c if the product of primes of b is greater than that of c in the ordering of the naturals.

Let ordering 3 be lexicographical such that (ordering 2,ordering 1)
.Then it should now be possible to have binary sequences b and c such that b > c in ordering 2 but b < c in ordering 1 hence b < c

Tomorrow will check whether this defines a well ordering of binary sequences...

Julius

16:20

Hi. Suppose that $X$ and $Y$ are discrete random variables with 1 and 2 as possible values. Then $E[X]=E[X|Y=1]P(Y=1)+E[X|Y=2]P(Y=2)$ but I am not given that $Y$ takes both 1 and 2 with nonzero probabilities. Can I then just say that $E[X|Y=i]=0$ if $P(Y=i)=0$? If so, why? I understand that it would be more involved if $Y$ were continuous, but it feels like in this case defining the conditional expectations as 0 should work somehow..

Maneesh Narayanan

16:38

2

Q: Let $f (z) $ be an entire function such that $|f (z)|≤K|z|$, $∀z∈\mathbb{C}$, for some $K>0$. If $f (1) =i$, then$f (i) $ is

Let $f (z) $ be an entire function such that $|f (z)|≤K|z|$, $∀z∈\mathbb{C}$, for some $K>0$. If $f (1) =i$, the value of $f (i) $ is (A) $ 1 $ (B)$-1$ (C) $i$ (D) $-i$ how can I able to solve this problem?totally stuck.

complex-analysis

@AkivaWeinberger here $f(z)=Az+B$ right? But I am not able to find the value $f(i)$. can you please explain?

user21312

Let x_1 * ... * x_n =1

How does one show that x_1 + ... + x_n >= n

?

Alessandro Codenotti

AM-GM

user21312

Thanks

Akiva Weinberger

17:01

@ManeeshNarayanan This is true for all $z$, not just for where $|z|>1$

In particular, it's true for $z=0$

so $|f(0)|\le K|0|=0$, meaning $f(0)=0$ and $B=0$

Maneesh Narayanan

17:24

ok. Thank you @AkivaWeinberger

BAYMAX

Hha steamyroot to tastyromeo

:)

I think if we jumple up steamyroot we would get tastyromeo @TastyRomeo right?

TastyRomeo

Yup

BAYMAX

I have a complicated looking qn

$\beta , h$ are fixed!

then is the following inequality true for all values of $\lambda$ ?

$1 + 4\lambda^4 sin^4(\frac{\beta h}{2} )- 4 \lambda^2 sin^2 \frac{\beta h}{2}+ \lambda^2 sin^2(\beta h) < 1$ ?

BAYMAX

17:55

this results to

Semiclassical

@BAYMAX as written, that's a bit difficult

BAYMAX

$\lambda ^2 sin^2 (\frac{\beta h}{2}) - 1 \leq 0$

Semiclassical

and actually it's trivially false if you take $\lambda=0$.

so presumably you mean $\lambda>0$

BAYMAX

i have to find a conditon on $\lambda$

Semiclassical

well

if $\lambda =0$, what happens to the inequality?

BAYMAX

17:57

for which the above inequality is true!

i meant $\leq 1$

there sorry!

Semiclassical

kk

since $\lambda$ is presumably real, we know that $\lambda^2\geq 0$

BAYMAX

yes

Semiclassical

and if $\lambda=0$ the inequality is in fact obviously true

so let's take that as settled and assume $\lambda \neq 0$

then $\lambda^2>0$, so we can simplify the inequality a bit

(subtract one from both sides, divide by lambda^2)

and you get $4\lambda^2 \sin^{4}(\beta h/2)-4\sin^2(\beta h/2)+\sin^2(\beta h)<0$

Transcript for

$Mathematics$ Mathematics