Mathematics - 2016-12-30 (page 3 of 6)

Secret

05:00

@TheGreatDuck So you want to construct a system of axioms suchthat you can nuke this?

> "if and only if a function is constant does it have a derivative of 0 for all real numbers"

that is, you want nonconstant functions whoose derivative everywhere is zero?

Null

but a function has to be defined at all of the elements in its domain

thats what i meant

DHMO

@Null so you meant point 1 instead of point 2

Null

yep, sorry

DHMO

yes, that is the definition of a function, then.

TheGreatDuck

@Secret um yeah... there are versions that still have a consistent system of calculus. It just changes the notion of antiderivative and derivative.

and a full set of understandable axioms would allow one to search for other alterable things that may produce intriguing results

I think of it like non-euclidean geometry

except instead of us thinking of it as geometry on a sphere

I think "could this other system do problems easier that we can convert back to the original problem"

for instance

that statement is fundamental in proving the following

"Any potential antiderivative must be continuous everywhere"

i could easily cook up examples where that is not true

honestly it would probably be better for me to define the axioms myself

but I know I'd probably end up doing something shoddy or contradicting

Secret

05:13

Well, initial analysis suggest if you nuke the axiom you mentioned above, (that is, using $f'=0$ for all $f \in \mathbb{C}(\mathbb{R})$), you pretty much nuked everything as when you go through the other axioms, and then remembering that 0 is a multiplicative absorber, then e.g. the chain rule and the addition law of derivatives will become zero for all $f$

GFauxPas

@dhmo how do you think we should present the graphs I made

Secret

so the derivative function becomes "multiplication by zero" effectively speaking

DHMO

@GFauxPas no idea

GFauxPas

I mean, I could just dump all four graphs there like I did on your discussion page, but I need to convey to the viewer what it is we're looking at. I tried adding a title over the graph but it doesn't look good when I do that

TheGreatDuck

@Secret chain rule only works for continuous functions. Also. I'm not saying it's 0 for all functions. Another possible negation would be the following:

"if and only if a function is piecewise-constant does it have a derivative of 0 for all real numbers"

the original axiom is now false

there exists some non-constant function whose derivative is 0

DHMO

05:18

@GFauxPas then add it under the graph?

TheGreatDuck

and it doesn't imply all functions have derivative of 0

because piecewise constant functions have no particular inverse by which to regain the identity function

there are other examples

the squaring function

periodic functions

etc.

Secret

So in your system, are ordinary constant functions $f=c$ are considered a subset of piecewise functions?

TheGreatDuck

well yeah

constant function are piecewise-constant

by definition of piecewise constant

Secret

ok so basically in ordinary deriviatives $f'$ will be 0 except for the jumps, while in your systems, even the jumps will be zero?

GFauxPas

DHMO might as well put on on the proofwiki page rather than on the image

so the graphs of $\operatorname{Arg}(W)$ and of $\operatorname{Arg}(\operatorname{Log})$ look very similar

Null

05:35

is there a term for discontinuous everywhere?

GFauxPas

So I decided to graph $|\operatorname{Arg}(W) - \operatorname{Arg}(\operatorname{Log})|$ to compare how similar they were

and weird stuff happens

TheGreatDuck

@Secret that's an example of one particular negation but yes.

the integral is a bit more interesting

@Secret not all systems would have to use that axiom. Another axiom could be "if and only if a function is periodic does it have a derivative of 0 for all real numbers"

Secret

Nuking the integral as a kinda of "inverse" of the derivatives (except for a constant) means nuking the fundemental theorem of calculus. That will mean your integral will be something that is not lebniz integral nor a riemannian integral
(possibly not any type of integral with a known measure, (I am not very good a measures however)). I am not sure what happens when you do that, but that will surely be an interesting case given how most of out integrals rely on its relation with the derivative in order to be evaluated

TheGreatDuck

well technically the antiderivative would still have a nice relation

we just wouldn't vary some constant c

it would be some function c(x)

i know in the piecewise-constant case that the actual integral pretty much always differs by some piecewise-constant function except in cases of divergence.

not 100% sure but I also have a strong feeling it's true for diff eqs. as well

i mean their solutions that is

not that that implies any means of actually finding the solution to a normal differential equation in regular calculus

that is arguably a much more difficult prospect

GFauxPas

TheGreatDuck

05:43

I thought of maybe calling these things non-Neutonian Calculus

Secret

Nuking (f+g)'=f'+g' will mean the lebniz product rule will be necessary if you want to recover some of the usual notions of derivative, unless you are happy to have that go to

TheGreatDuck

but that's a bit contrived seeing as how many many men created calculus.

GFauxPas

$||\operatorname{Arg}(\operatorname{Log}(z))| - |\operatorname{Arg}(W(z))||$

I guess they're not similar after all

TheGreatDuck

@Secret not interested in nuking that rule. Only interested in nuking the one I mentioned. However, if the others prove fruitful in certain ways then by all means I would examine them.

GFauxPas

oh it's late

I need to sleep

good night

TheGreatDuck

05:47

another potential axiom that could be useful is:

if for all real numbers c' = 0 then (c*f)' = cf

less strong then the product rule

but the product rule could create a contradiction

(for instance, if the derivative of x^2 is 0 by the constant axiom)

then (xx) ' = 2xx'

however

this implies either x is 0 everywhere (false)

or x' is 0 everywhere

Null

x is 0 everywhere i like :D

TheGreatDuck

which is not supported since x cannot be expressed as f(x^2)

@Null To put it in context, me and secret are discussing the prospect of alternate calculus axioms.

Null

yep

just found it funny ;)

TheGreatDuck

ah

Secret

@TheGreatDuck Well, consider c=1, then $c'=0$. Therefore by the cf axiom , (1f)'=1f=f, meaning all derivatives of f is itself

TheGreatDuck

05:51

oops

I meant = cf'

if for all real numbers c' = 0 then (c*f)' = cf'

doesn't seem to imply that any falsehood or contradiction with the postulate I proposed

and doesn't seem explicitly provable.

Secret

that can be dervied using $1'=0$ and $(f+g)'=f'+g'$
Set g=0, f=cf. Then (cf+0)'=cf'+0'=cf'+0=cf'

TheGreatDuck

granted, there's also the strong possibility and suspicion I have that alternate derivatives like this will act different depending on the expression of a function rather than the values of a function.

which may or may not be bad.

Secret

The periodic $f(x+a)'=0$ case I am still checking. IT seems interesting

TheGreatDuck

yeah it does

i only recently learned of it

finding it's relation back to regular calculus is an intriguing prospect.

Secret

although that will mean fourier transforms and foruier series will nto be applicable to thse fucntions else their derivatives to be identically zero

that's not really a problem, because there are functions canot be handled by fourier series anyway

TheGreatDuck

05:55

shrugs I don't know how far these things can go.

and I think their usefulness (while potentially analytical in some way) I believe will only extend to being usable in solving differential equations more efficiently.

Secret

Naively speaking, $f(x+a)'=0$ might be useful in checking whether a function contains an oscillatory component, but details I still need to check

TheGreatDuck

for instance the equation $y'' + y = d$ has the general solution $y = c1sin(x) + c2cos(x) + d$

replace d with a version of the heavyside function as in the aformentioned alternate system and you've pretty much solved that differential equation

just have to ensure that the two independent solutions have piecewise constant cn such that they are continuous

@Secret interesting. I just noticed it was a set of functions that fulfilled one of my requirements for having the same solutions to differential equations as usual differential equations.... without rendering the equation meaningless I mean.

cause we make assumptions based on c_n always being the same function type

and then add them subtract them etc while still assuming they are still a ____ function

periodic happens to be one of four known sets that fulfill that

Secret

The periodic rule also covers the constant case, since constants are also a periodic function (a wave of infinite wavelength)

TheGreatDuck

well not neccessarily

the other sets I know of fulfilling that rule I specified are:

constant (obviously)

periodic function

piecewise constant functions

and all functions (cause I would hope there are not two functions that when composed are no longer a function)

@Secret well while examining them is intriguing and all. My prospect is to try and develop a set of postulates including that one that truly govern real number calculus. I suspect that the answer might be really trivial, but I don't want to choose a cop out.

for instance

I could get away with two postulates if I make the second one:

"if there exists a real number x such that the derivative of f at x is not zero, then the derivative of f is (insert standard limit definition here)"

that feels like a silly cop out though

granted, it would put more emphasis on my statement as in "we only manipulate this one".

then again that would mean x*floor(x) would have a normal derivative rather than an altered derivative of floor(x)

so then again after thinking it through... it's not that trivial.

DHMO

@Secret you meant infinitesimal

@TheGreatDuck so you're creating your own rules now?

TheGreatDuck

06:09

@DHMO we're discussing negations of calculus axioms. specifically of

"if and only if a function is constant does it have a derivative of 0 for all real numbers"

DHMO

this isn't an axiom at all.

TheGreatDuck

it can be.

DHMO

but if you negate this theorem, your system would be inconsistent

unless you negate every single axiom that leads to that theorem

TheGreatDuck

no

only if you use the limit definition of the derivative

DHMO

so you're negating other axioms that lead to this theorem

TheGreatDuck

06:11

in this situation, the axioms define the derivative and lead to the limit formula.

DHMO

so what's your new definition of the derivative?

Secret

I think the explicit form does not matter, because it can be defined by the axioms

TheGreatDuck

@DHMO That's what we're trying to figure out. I'm trying to figure out a clear set of independent axiomatic statements including that axiom that fully define real number differentiation and antidifferentiation

the integral is it's own thing so it can be left to the side as riemann sums or whatever.

DHMO

@Secret are you interested to discuss the transcendence of e?

@TheGreatDuck what is differentiation?

and why do you think that the current system is inadequate?

Secret

@DHMO Let me tried a bit more on redoing thegreatduck's derivation based on the MO link (should not take too long) and I will attend you shortly (earlier this morning I have been teaching chemistry in the h bar)

TheGreatDuck

06:15

@DHMO I don't think the current system is inadequate. I am interested in looking at systems using negations of that axiom and so in order to do that one should first have a clear set of rules so that if you do negate it everything is still a clear system of calculus.

though I definitely wouldn't claim it to be an equivalent calculus. :p

DHMO

@TheGreatDuck that is not an axiom.

TheGreatDuck

@DHMO it can be with no problems

because

the derivative need not be defined explicitly via the limit formula

one might argue that is derived via the axioms

DHMO

it doesn't make it an axiom

TheGreatDuck

i never said it was commonly accepted as such

but if I choose to use it as an axiom what is to stop me?

DHMO

the entire mathematical community.

TheGreatDuck

06:17

it is a true statement.

and it's only proof comes from the limit definition

which means reversing it and claiming the latter proves the former might be odd but certainly not invalid logic.

DHMO

alright, good luck inventing your own calculus

TheGreatDuck

@DHMO you're being a jerk, do you know that?

also

there are axioms of calculus

see here

13

Q: Independence of Leibniz rule and locality from other properties of the derivative?

The following is meant to be an axiomatization of differential calculus of a single variable. To avoid complications, let's say that $f$, $g$, $f'$, and $g'$ are smooth functions from $\mathbb{R}$ to $\mathbb{R}$ ("smooth" being defined by the usual Cauchy-Weierstrass definition of the derivative...

DHMO

@TheGreatDuck I apologize.

TheGreatDuck

trying to turn it into an equivalent set including the statement I was interested in is not at all invalid thinking.

I might be working purely in real number calculus

but that's just my particular knowledge level

DHMO

so if you negate that axiom

what do you get?

TheGreatDuck

06:21

one particular version that is a negation would be

"if and only if a function is piecewise-constant does it have a derivative of 0 for all real numbers"

basically, there exists some non-constant function whose derivative is 0

DHMO

even at the discontinuous point?

TheGreatDuck

yes

all real numbers

DHMO

what is the derivative of |x| at x=0?

TheGreatDuck

normally, undefined.

DHMO

under your calculus?

TheGreatDuck

06:23

give me a second to think about it. I have to rewrite it better for the other version.

DHMO

I fail to see how the 6 axioms in your link above can lead to x' = 1

TheGreatDuck

note: they are for abstract differentiation

might not be purely numerical

2*[x>=0]*x - x = |x| and has a derivative of 2x - 1 which means at 0 it has a value of -1

I see your point

it could also be 1 if I write it a different way.

granted, that simply means the derivative in this context isn't an operator

it's a function upon the set of expressions

which is what I said to secret earlier as a problem

DHMO

You might be interested in Cantor function:

Cantor function

In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous. It is also referred to as the Cantor ternary function, the Lebesgue function, Lebesgue's singular function, the Cantor-Vitali function, the Devil's staircase, the Cantor staircase function, and the Cantor-Lebesgue function. Georg Cantor (1884) introduced the Cantor function and mentioned that Scheeffer pointed out that it was a counterexample to an extension of the fundamental theorem of calculus claimed by Harnack. The Cantor function was discussed and popularized by Scheeffer (1884...

under your system, the derivative would be 0 everywhere in (0..1) while the value goes from 0 to 1

TheGreatDuck

yup

but the antiderivative is x*c(x)

DHMO

what is c?

TheGreatDuck

06:28

and I can say that there is some piecewise constant function d(x) such that x*c(x) + d(x) is the integral of the cantor function

c is the cantor function

Secret

Using this as a template:
Z. $\exists f : f'\ne 0$

U. $1'=0$

A. $(f+g)'=f'+g'$

C. $(g \circ f)'=(g'\circ f)f'$

L. The value of $f'(x)$ is determined by knowing $f$ in any neighborhood of $x$.

Now replace Z by P:
P: $\exists f : f(x+a)=f(x), f'= 0$
Then:

U = the same U
A = the same A
C = if $f$ periodic, then $(f^n)'=0$
Lebniz: if $f$ periodic, then $(fg)' = 0$ for all $g$
L = same L

TheGreatDuck

I'm not sure if L is explicitly true anymore.

Secret

Well as long restriction can be treated as composition, L will remain true

DHMO

I don't understand L.

TheGreatDuck

as a function being periodic is not determinable by looking at the neighborhood of x

Secret

06:31

although I am not very sure how the periodic case will screw it up given all periodic functions and their products and powers have zero derivatives now

TheGreatDuck

@DHMO basically, the derivative is always based on the values around a point

Secret

So what you proposed is actually a subset of the usual differential algebra that we are familar of

TheGreatDuck

not the entire function

@Secret in a weird twisted way, yes.

Secret

It could be useful if your diff eqt have periodic solutions, I guess

TheGreatDuck

except it's not operational derivation

it's actually symbolic differentiation

Secret

06:32

for example take: y"+y=d
If y periodic, then y"=0, thus you end up with a boundary condition y(x+a)=d

TheGreatDuck

as @DHMO pointed out, the values may change depending on how you write a function

@Secret intriguing.

Secret

So, I guess, your subset of differential algebra will be useful in testing whether periodic solutions solve the differential equation, however more is need to be check to be sure

TheGreatDuck

well...

Secret

In particular, the integral will now have an arbitrary periodic function

TheGreatDuck

it's more like

if d is periodic than the solution will be this thing: c1*sin(x) + c2*sin(x) + d

and then it's a matter of finding whatever c's somehow give the correct solution

granted

i have no clue how to do that

there's probably a specific subset of functions governing that

but it's not periodic

best I can do is find a continuous solution

but it wouldn't be right

shrugs. it's an interesting concept. Not as trivial as the piecewise constant one.

that one is easy to convert back

just choose piecewise constant c's so that y is continuous

boom

yer done

I'm going to go to bed

DHMO

06:36

every periodic function has derivative zero?

TheGreatDuck

in this system yes

Secret

and, as derived from the axioms, every powers and products of them

TheGreatDuck

@Secret wanna reconvene later?

DHMO

interesting

Secret

sure

TheGreatDuck

06:37

@Secret ideally any mutlivariate composition

because believe it or not

for periodic f and g

DHMO

so thomae function has derivative 0?

TheGreatDuck

h(f,g) is periodic

for all h

DHMO

@TheGreatDuck cos(x) is periodic. cos(x*sqrt(2)) is periodic.

TheGreatDuck

@DHMO the cantor function is 0 in the piecewise constant system.

DHMO

their sum is not.

@TheGreatDuck i'm talking about thomae not cantor

Secret

06:39

$h(f,g)'=h'(f,g)(f'+g')=0$, hence h is periodic

DHMO

@Secret cos(x) is periodic. cos(x*sqrt(2)) is periodic. their sum is not.

TheGreatDuck

@DHMO see here math.stackexchange.com/questions/1992503/…

DHMO

@Secret periodic implies zero derivative. its converse is not assumed?

TheGreatDuck

and here

-4

Q: A composition of a periodic function that is somehow not periodic.

I have this conjecture: $$\forall f: R \to R(\exists g:R \to R (x \cdot f(x) = \int f(x) dx + g(f(x))))$$ Basically, my problem is that I am very very sure this is true. I have no way to prove it, but I strongly believe it to be true. However, $f(x) = (x \mod 1)$ seems like a counter-example. T...

integration periodic-functions

unless i misunderstood

the answers I recieved must be wrong by your counter-example.

Secret

@DHMO In his system, things have zero derivative only if they are periodic (including constant functions)

TheGreatDuck

06:41

@DHMO it is if and only if

DHMO

then your system is inconsistent, as my counter-example suggests.

TheGreatDuck

must be rationally periodic then

idk

Secret

@DHMO Due to $(f'+g')=f'+g'$, your example will be considered periodic to his derivative, which is...kinda... idk

TheGreatDuck

something about that in the answer i recieved in that link

@Secret we're not distorting the definition of periodic

:p

Secret

In that case, you need to find a way to take account of DHMO's counterexample

DHMO

06:42

1 min ago, by TheGreatDuck

must be rationally periodic then

TheGreatDuck

see the answers i linked

they probably explain the issue

it's what I based by h(f,g) statement on

either that or they are wrong answers

i gotta go

Secret

There is still no algebra people yet. Let me just do some final check on a table and I will head to transcendetal number room. Also if some of those who visited my room have questions, please ask there, it is getting too inactive to maintain the room

robjohn

07:00

@Secret Sorry I'm not an algebra person.

Secret

Yup. I know

Null

is $\mathbb{R}\setminus\mathbb{Q}$ with "holes" like $\mathbb{N}$?

discrete is the word i think

Perturbative

Are single sets equivalent sets?

DHMO

@Null what does discrete mean?

no it isn't discrete

Secret

07:45

[Division by zero] Everyone said not to divide by zero is because they **** up all rules. Turns out, they are perhaps the most rule abidding (for associative division by zero algebras) of all:

$0^2$ determines the fate of the zero terms by associativity, zero inverse determines the fate of all products of it

and then all zero terms (including the zero inverse) dictate the fate of 3n entries in the addition structure by one sided distributivity

It is PRECISELY because they are so rule abiding is why they are such a headache

(because having $3n$ out of $n^2$ entries knocked out, the table is going to have a sizable portion to become a one sided null semigroup)

How to prove this for infinite structures is a work in progress

Basically, there is very little wiggle room for associative division by zero algebras (the number of constrained entries in the addition sturcture is roughly halved by having only one sided multilpicative identities) until it gets large enough

to the point that for $n=3$ it reduces to $\mathbb{Z}/3$

Secret

Anyway, wrong timing, i should head to the transcendental room now...

Dair

as someone who isn't that good at algebra, is a zero algebra just an algebra that contains 0 * r = 0 and 0 + r = r?

Secret

I think you mean zero term algebra. I don't recall there is anything named zero algebra. However there is something called a zero object

Dair

"for associative division by zero algebras"?

Secret

O, that...

Alessandro Codenotti

07:57

@Secret what do you mean?

Balarka Sen

do you really want to know

Alessandro Codenotti

Wait, I replied to the wrong message, stupid phone

@Perturbative what do you mean?

Dair

@BalarkaSen Maybe I don't haha.

Alessandro Codenotti

Nope @Balarka, I was trying to answer to @Perturbative!

Secret

it basically means any associative algebra that allows a multiplicative inverse of zero

This is possible by knocking out $0*n=0$ and anything that lead to it

Wildcard

07:59

This might sound like a troll, but it's a serious question:

Dair

what do you mean by "knocking out $0 \cdot n = 0$"?

Wildcard

I wonder how many mathematicians I am likely to meet who are capable of recognizing the degree of precision and variability contained in the English language—not as something which should be clipped carefully to disallow any ambiguity, but as something which has greater expressive power for variability of ideas than any other mathematics yet invented?

Secret

@Dair Perhaps it will be better to discuss that in the room now that the chat start getting busy

Dair

08:25

hey @Daminark

DHMO

Let $k = \displaystyle \sum_{n=1}^\infty \frac 1 {2^{n^2}}$.

What do we know about $k$?

SoumyoB

we know it's a real number

and we know it's less than 1

and it's positive

@DHMO

DHMO

that's very useful thanks

can we prove/disprove that it is transcendental?

Balarka Sen

08:51

lol

5

Secret

Sullivan invented it on the spot. — András Szűcs Oct 15 '14 at 11:44

always cool to see these

Secret

09:11

akiva, maybe you can help me on this semigroup question?

0

Q: Necessary and sufficient condition to show the surjectivity of an action if the existence of right inverse is not known

Given a semigroup $S$ with possibly infinite elements , it is known that there is an element $a$ such that its left semigroup action on all other elements $x \in S$ either fixes it or map to a distinct element i.e. $ax=x$ or $ax=b$ where $b \in S$ and $b \neq x$. It is also known that a left inve...

permutations semigroups

Mike Miller

09:30

@BalarkaSen That's a nice desciption.

Secret

hmm... asymptotes, that's easy for uncountable structures with some kind of topology, but I wonder how can we generalise that to countable systems... hmmm...

so for surjectivity, we need something more than having the image of each point being not the same as the point

whatever that means, I 'll sort that out when I finsih my transcendental revision

Balarka Sen

@MikeMiller It's not obvious to me why a Lie group is always supposed to contain a circle subgroup though.

Well, I suppose compact ones, because that's obviously false otherwise

Mike Miller

You're not trying to null-bord the noncompact ones.

Alessandro Codenotti

@Balarka is there an hat for offering a bounty?

Balarka Sen

@Alessandro IIRC on the last day of winterbash

Secret

09:47

Nov 7 at 3:42, by Brody

@Kaumudi Range cannot be determined analytically, for now you need the graph of the function, which will tell all

suprisingly relevant

seems it holds for semigroup actions as well

Wonder if groups will fare better...

DHMO

10:02

@AlessandroCodenotti can you determine if sum 2^(-n^2) is transcendental?

i would be quite surprised if it is algebraic

never mind

21

A: Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results?

I have checked with Introduction to Algebraic Independence Theory, where it is mentioned in the preface (p. V) that D. Bertrand and independently D. Duverney, Ke. Nishioka, Ku. Nishioka, I. Shiokawa (DNNS) deduced results on algebraic independence of the values of theta-functions at algebrai...

Secret

11:11

in Algebraic/Transcendence Theory, 2 days ago, by DHMO

so $f^{(k)}(j) = 0$

Why are prime numbers so confusing

The Substitute

Off topic: Is $n$ choose $k$ defined for $k$ not an integer?

Ali Caglayan

@TheSubstitute just use gamma function instead of n! in definition

The Substitute

@AliCaglayan Thanks that's what I thought. For some reason I was getting different answers in mathematica when I rewrote the binomial coefficient this way.

Akiva Weinberger

@TheSubstitute Remember that $x!=\Gamma(x+1)$, not $\Gamma(x)$.

The Substitute

@AkivaWeinberger Oops! Thank you for that.

Akiva Weinberger

11:22

Also, if $n$ isn't an integer but $k$ is (the reverse from what you have), you don't need Gamma to define $\binom nk$.

It'd be $n(n-1)(n-2)\dotsb(n-k+1)/k!$.

Ali Caglayan

@AkivaWeinberger it seems k is not the integer in this case

hello @Alessandro

Alessandro Codenotti

Hi

Studentmath

I've got a small question, I am assuming $7^{1/3}\in Q(5^{1/3})$. Showing that leads to $Q(7^{1/3})=Q(5^{1/3})$, I showed as required (with the basis $(1,7^{1/3},7^{2/3})$ that there are $a,b,c\in Q$ so that $a+b7^{1/3}+c7^{2/3}=5^{1/3}$. Now they want me to show, using the transformation $Ta=5^{1/3}*a$ that $a=bc=0$.

I don't get it - I derived using two different basis for the transformation that $(t-5^{1/3})^3=t^3-5$ (the contradiction yells here). I can follow suit placing $t=1$ and replacing $5^{1/3}$ with the previous equation and show that a=b=c=0. Which makes me thing I am missing something/am wrong

Since the question clearly asks to derive something else..

Anyone can see where I am skipping a step?

Ali Caglayan

11:38

@Studentmath If you can show a=b=c=0 then can't you show a=bc=0?

oh wait

If $Ta=5^{1/3}a$

then $Ta+T(b7^{1/3}+c7^{2/3})=T(5^{1/3})$

Actually if you apply T twice

on the left hand side you get some mess

but on the right you get an element of Q

Secret

I wonder what does a polynomial look like graphially before expanding it...?

Ali Caglayan

so by equating components you have what you need I belive

Secret

in particular, geomety wise, this theorem always seemed like magic to me:

Cayley–Hamilton theorem

In linear algebra, the Cayley–Hamilton theorem (named after the mathematicians Arthur Cayley and William Rowan Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation. If A is a given n×n matrix and In is the n×n identity matrix, then the characteristic polynomial of A is defined as p ( λ ) = det ( λ I n − A ) , ...

Studentmath

11:54

@AliCaglayan Yes, of course I also have a=bc=0, but it makes me think I am missing something along the way. And thanks! I will try to look at that out. Generally I am supposed to conclude that by looking at the polynomial of T under two different basis (which is what yield $(t-5^{1/3})^3=t^3-5$)

DHMO

the functions R->R with rational periods form field under pointwise addition and multiplication

Alessandro Codenotti

I'm not aure about the general case but I think that if you consider matrices over a field you can think about the matrix ring as an extension of the scalar matrices field and use the fact that the minimal polynomial divides the characteristic one @secret

I'm not sure about the proof in the general case but I think that if you consider matrices over a field you can think about the matrix ring as an extension of the scalar matrices field and use the fact that the minimal polynomial divides the characteristic one @secret

Secret

12:09

Well yes that's how we proved that back in linear algebra class. What I am more interested in is that since linear operators represent transformations in a vector space, and then a matrix polynomial is some map from the vector space to itself (thus some kind of nonlinear transformation), what is the geometric meaning for a matrix that solves the polynomial (and hence vanishes by cayley hamiltonia theorem)?

that is what does it mean geometrically in terms of transformations in vector spaces when a matrix solves the polynomail and hence vanishes. Or rather, what does a root of a matrix polynomail mean geoemtrically?

Alessandro Codenotti

12:24

@DHMO are you sure about the multiplicative inverses?

Kaj Hansen

12:39

@TedShifrin, I think of you whenever I see this: gifbin.com/bin/122014/….

DHMO

@AlessandroCodenotti quite.

except the functions that contain zero, right

they form ring then

Akiva Weinberger

Maybe do the one-point conpactificstion of $\Bbb R$ instead

aka the one-dimensional projective plane

$\Bbb R\cup\{\infty\}$

Mats Granvik

12:54

Is the Fibonacci sequence deterministic or non-deterministic?

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$Mathematics$ Mathematics