Mathematics - 2016-11-18 (page 3 of 4)

Ramanujan

13:33

If anyone get then tell me(or any link where can I get it's proof)

DHMO

@Ramanujan use induction and the formula $\sin(2\theta)=2\sin\theta\cos\theta$

Null

@DHMO @DHMO yes, i meant f(f)

DHMO

alright

Ramanujan

@DHMO any proof? Iam not trying check,trying to proof

Null

eh

DHMO

13:37

@Ramanujan the proof is induction by the formula $\sin(2\theta)=2\sin\theta\cos\theta$

Null

trying to proof doesn't include getting it spoonfed. altho it might help to read similar proofs. (to get the identities straight)

as DHMO said, induct your way through it

Ramanujan

@DHMO wait a minute

23rd one

DHMO

@Ramanujan use the identity you just proved

Ramanujan

Before seeing hint I don't know that identity

Null

actually the identity is a little bit different

DHMO

13:42

then multiply $\sin\left(\dfrac{2\pi}{2^{64}-1}\right)$ to the whole expression

Null

powers of 2, not simply n

$2^n$

DHMO

@Null ?

it is $2^n$ in both instances

Null

i dont see here powers of 2

DHMO

@Null I see clearly $\cos(2^n\theta)$

Null

oh well

teadawg1337

13:44

The notation is poor, but it's the same problem

Null

yes

DHMO

@teadawg1337 which notation is poor?

Ramanujan

@DHMO do you mean multiply and divide or multiply on both sides?

DHMO

@Ramanujan whichever you like

teadawg1337

@DHMO The lack of parentheses in the original image

DHMO

13:45

@teadawg1337 I see

cfp

@Semiclassical Just noticed the bounty you placed on my question. Thanks again, it's much appreciated.

Ramanujan

@DHMO and then multiply and divide by 2?

DHMO

@Ramanujan sure

teadawg1337

Oh dear, I just noticed that the textbook being used has the same notation issues

Null

How can the Cauchy-Schwartz inequality be used on sequences of numbers that might not have "enough" elements?

DHMO

13:49

@Null for example?

Ramanujan

@DHMO iam left with this…then?

DHMO

@Ramanujan it should be $2^{64}$ not $2$

Ramanujan

@DHMO in denominator?

DHMO

yes

Null

for $a_i$ we take the numbers 1 to 10. For $b_j$ the numbers 20 to 50. Would that be simply not intended? @DHMO

(and n= higher then 10)

DHMO

13:51

@Null then the inequality does not apply?

you can also fill the other sequence with zeroes

Ramanujan

DHMO

where did the denominator of the first line come from?

Ramanujan

I will show full Sol

@DHMO this iam getting

@DHMO are you there?

DHMO

@Ramanujan I am not getting the first step

Vrouvrou

Hello omeone know the Lusternick-Schnirelman theory ?

Ramanujan

14:07

@DHMO for each angle of cos I multiplied and divided with sin of that angle ,and then to get in 2sinacosa form I multiplied by 2^(no.of terms)

DHMO

@Ramanujan fair enough

now you can continue by using $\sin(\pi-x)=\sin x$

Lusternik–Schnirelmann theorem

In mathematics, the Lusternik–Schnirelmann theorem, aka Lusternik–Schnirelmann–Borsuk theorem or LSB theorem, says as follows. If the sphere Sn is covered by n + 1 open sets, then one of these sets contains a pair (x, −x) of antipodal points. It is named after Lazar Lyusternik and Lev Schnirelmann, who published it in 1930. == Equivalent results == There are several fixed-point theorems which come in three equivalent variants: an algebraic topology variant, a combinatorial variant and a set-covering variant. Each variant can be proved separately using totally different arguments, but each variant...

Ramanujan

@DHMO do I need to know the value of 2^64 - 1?

DHMO

@Ramanujan no

Ramanujan

Then how?

DHMO

1 min ago, by DHMO

now you can continue by using $\sin(\pi-x)=\sin x$

Vrouvrou

14:10

@DHMO i want to know if we can apply this theory on ordinary differential equation ?

DHMO

@Vrouvrou no idea

Ramanujan

@DHMO I can't find any term of sin(π-x)

Balarka Sen

@Vrouvrou How do you plan to apply this to ODE's?

DHMO

@Ramanujan use $\sin x = \sin (\pi-x)$

Balarka Sen

@DHMO the LS theory is not quite limited to that theorem

DHMO

14:12

@BalarkaSen I'll shut up then

Vrouvrou

@BalarkaSen i read many papers about this theory and all apply it on pde , i want to know if it can be apply it on ode

Balarka Sen

I have never heard it being applied to PDE's. By LS-theory I presume you meant the study of LS-category.

@DHMO NB I don't know much about it either

Vrouvrou

@BalarkaSen yes they obtain cat(\Omega) of solutions

Ramanujan

@DHMO it's not getting reduced

DHMO

@Ramanujan apply it only to one of the sines

no

use $\sin x = - \sin(2\pi-x)$

teadawg1337

14:17

Instead of using that identity, I believe it would be easier to observe the period of $\sin(\theta)$. The inclusion of $\pi$ in the problem isn't a coincidence

Danu

@BalarkaSen Heyo

Null

many people here

Balarka Sen

Hi @Danu.

Still complex geometrizing hard?

Danu

Yeah man

I'm almost done!!!

9 pages 'til I finish chapter 5 of Huybrechts

All the big theorems rolling in now

it's great :D

Balarka Sen

yay

Danu

14:19

All the $H^q(\Bbb C\rm P^n,\mathcal O(k))$, done :D

This Kodaira vanishing theorem is excellent :D

Ramanujan

Numerator

Null

how is this any different from cryptology haha

Balarka Sen

@Danu Yeah, I vaguely remember it

DHMO

@Ramanujan nice

Danu

@BalarkaSen Now I'm doing that "Weak Lefschetz Theorem" you mentioned a while back.

Ramanujan

14:21

Then?

Balarka Sen

Ah I see

Danu

(relating cohomology of a hypersurface to that of total space)

DHMO

@Ramanujan the you can cancel

Danu

What proof did you see?

This is all SES's of sheaves

Balarka Sen

I know there's a Morse theory proof, and if I keep reading Milnor I'll see it.

Danu

14:22

oh, really?

He talks about complex geometry?

Balarka Sen

yup.

Danu

I really wanna know about Morse theory.

Ramanujan

@DHMO really thanks for help,i got now,so I think we can prove that formula in the same way,again thanks

Balarka Sen

it's cool stuff; i started reading some for passing time

DHMO

@Ramanujan you are welcome

Vrouvrou

14:24

@BalarkaSen please do you know what is the LS category of an interval ?

Balarka Sen

@Vrouvrou you should be able to answer that yourself if you know the defn

Ramanujan

@DHMO btw how you got that we need to use sinx=-sin(2π-x)?

DHMO

@Ramanujan because $2 \times 2^{64} - 2 = 2(2^{64}-1)$

Vrouvrou

@BalarkaSen i think tht is 1

Balarka Sen

it's 0. the ls-category is 1 less than the number of open sets you cover by

Null

14:31

how is a piecewise function called that only fills in the gaps? like for (sin(x)/x), y=1 for x=0?

(point gaps i mean)

Akiva Weinberger

So in physics class we were discussing the lift force, $F_L$, and specifically its $y$-component $F_{Ly}$

I think it's great that the force that makes planes fly is called $F_{Ly}$

Balarka Sen

:P

Null

@AkivaWeinberger no, you are kidding now? :s

Ramanujan

64/16 just cancel 6 you are left with 4/1

Akiva Weinberger

Hopefully $F_{Ly}=-mg$ or you're gonna have a bad time

(or you're landing)

teadawg1337

14:35

@Ramanujan Woah hold on, where are you getting 64/16?

Null

@DHMO why in particular is sin(36) good to know?

Akiva Weinberger

@teadawg1337 His brain?

DHMO

@Null because it's nice

teadawg1337

@Ramanujan That's also not how you simplify fractions...

Balarka Sen

@Akiva "$m$y $g$od [and more screams]"

Ramanujan

14:36

@teadawg1337 I read that somewhere

arctic tern

@teadawg1337 whoosh

Null

@DHMO well -1 roughly will now be in my head, congratz on that :P

teadawg1337

I thought it was in reference to the problem that was being worked on previously

Akiva Weinberger

@teadawg1337 It's a sort of mathematical joke — $\dfrac{\cancel64}{1\cancel6}=\dfrac41$ is the wrong way to simplify fractions but gives you the right answer anyway

(LaTeX help? How to do diagonal strikethroughs?)

Ramanujan

@teadawg1337 hah,that problem was solved(over)

Vrouvrou

14:38

@BalarkaSen i have this definition : If $Y$ is closed subset of a topological space X , the LS category is the least number of closed and contractible sets in X which cover Y

teadawg1337

I'm aware of the joke, I just didn't realize it was a standalone statement at first

Akiva Weinberger

TeX: {
extensions: ["cancel.js"]
} $\cancel6$

Aw, didn't work

Secret

How most people research maths:
1. Encountered an or a bunch of interesting problem
2. identify a natural structure in the problem
3. Describe the structure with axioms
4. Prove and derive theorems
5. When the framework is established, generalise or abstract to similar objects
6. Rinse and repeat

How I research maths:
1. Choose a structure or mathematical object
2. Generalise it by discarding axioms
3. Use this new set of axioms to construct a new object
4. Make sure the constructed object exists and nontrivial

Balarka Sen

@Vrouvrou Ok. That's not the standard definition, which is 1 less than the number of open/closed sets.

DHMO

@Secret I prefer the bottom way lol

Vrouvrou

14:42

then if i use this definition the LS category of an interval is 1 @BalarkaSen

Secret

Both approaches are important. The top approach is important when solving real world problems. The bottom approach is more like pure maths/fine arts approach and focus on building as many of these structures and then describe them later.

The bottom approach is more experimental oriented, possibly because of my physics background

Balarka Sen

yes

Secret

Most of the time when I researching on things, I am usualyl drawn to how instrinsically interesting they are. To me, applications are often just an excuse of making them more tangible so I can "mess with it with my hands"

Vrouvrou

@BalarkaSen thank you

Null

15:09

@Secret so you are lego, but others playmobile?

Secret

possibly, although I can switch between the two modes depending on the problem in question

Astyx

Hi chat

user243301

Good weekend all users! Bye.

Mike Miller

My argument appears to be invalid. Hopefully I can fix it later.

DHMO

@Astyx hi

Astyx

15:29

How would one prove that if $f:\Bbb R_{+}\to \Bbb R$ satisfies $$f(nx)\to_{n\to +\infty}0$$ for all $x\ge 0$, then $$f\to_{x\to +\infty}0$$

Mike Miller

@Secret That is not how most people research math. Certainly the word 'axiom' doesn't show up in anything I've ever written, or am liable to.

Secret

In that case there is a lot more to learn cause I am only familiar with one of the ways to study abstract algebra (possibly how I made that mistaken claim above).

Mike Miller

I do not study abstract algebra :)

Secret

How do researchers study topology?

Mike Miller

(Most people in algebra study things like algebraic geometry or commutative algebra, where a lot of the time the interest isn't in the structure of some axiomatized object, but some more geometric question)

Secret

15:36

i see

s.harp

@MikeMiller This is an example of a self-contradictory statement

Mike Miller

"any papers" :P

Balarka Sen

I was imagining if Akiva would say that.

Mike Miller

@Secret Sorry for not responding, I intend to. But I'm a little busy this morning.

Secret

it's ok, take your time. I am also in the middle of something

Akiva Weinberger

15:42

@BalarkaSen Say what?

Ohh

He didn't say 'axiom', he said ''axiom'', doesn't count

Mike Miller

@Secret So I think it's impossible to say something about how "topologists" work. I can say something about how I work. I have questions that I think are interesting (which 3-manifolds embed in $\Bbb R^4$? is one of my favorites!), and there is a broad spectrum of tools which have historically been developed. So a lot of my thinking is either towards extending those tools to be more useful, or finding ways to use them. Sometimes I have a flash of insight and think

"Ah, I can use (idea) to try to tackle (question)", and think about that for a while, and then it doesn't work.

2

Secret

Hmm, it sounds kinda similar to organic synthesis in chemistry, where the interesting thing is a target molecule to be synthesized. There are theories and reaction pathways historically developed, and our job is to optimise these pathways in order to make the molecule more efficiently

Therefore to me, it seems like the maths tools will be used to figure out which 3 manifolds will embed in $\mathbb{R}^4$ or to better characterise them

Astyx

15:59

Has one of you studied in France ?

Akiva Weinberger

And were one really genius they could stumble upon an altogether new tool

abenthy

16:11

Is any path-connected space $X$ with $\pi_n(X) = 0$ for $n > 1$ homotopy equivalent to a CW-complex?

Balarka Sen

Yes.

Actually, um.

abenthy

I actually don't think it is.

Balarka Sen

Look at say the Warsaw circle. I believe that has higher homotopy groups all $0$ just fine.

Mike Miller

weakly contractible does not imply contractible

Balarka Sen

how is that relevant?

Mike Miller

16:23

what happens if a weakly contractible space is homotopy equivalent to a CW complex?

Balarka Sen

oh, I thought that was a reply to my example. yes, it's contractible.

Gridley Quayle

Any hints? f(t) is continuous and t^-2/3 is continuous on (0,1] so if we take the limit from the right the product should be continuous on [a,1] but I am not sure how to formalise this

abenthy

@BalarkaSen Thank you. This is a good counterexample. :)

Btw. when reading about mostow's rigidity theorem, nobody mentions the connectedness. Is it usual for people in this subject to only consider connected manifolds?

Mike Miller

yes

abenthy

So mostow's rigidity theorem only holds for connected manifolds?

Mike Miller

16:33

but a proper definition of hyperbolic is "has universal cover $\Bbb H^n$", anyway

abenthy

oh ok

Mike Miller

i mean, for disconnected manifolds you just need to check componentwise

abenthy

isnt it a riemannian manifolds with constant sec. curvature -1?

Oh okay. I see. So it does hold for disconnected manifolds :)

Mike Miller

when $M$ is connected, the two statements are equivalent

there's just never any interesting reason to consider disconnected things when one can always say "check components"

abenthy

makes sense, thank you

Balarka Sen

16:37

@GridleyQuayle Since $f$ is continuous on $[0, 1]$, you can bound it by a constant (why?). Work with that.

DHMO

@BalarkaSen wow, nice

Gridley Quayle

f is bounded, by M say, by the Extreme value theorem. hence $ t^{-2/3} f(t) \leq Mt^{-2/3} $ Integrating we see that the RHS converges to 3M (finite) thus the LHS converges.

Thanks @BalarkaSen

Balarka Sen

You have proved the limit does not blow up to infinity. There's still an issue about showing it actually exists.

Akiva Weinberger

I suppose it's theoretically possible that the integral has a weird oscillation thing as $a\to0$

DHMO

@BalarkaSen let the boundary be $[m,n]$.
$m \le f(t) \le n$
$mt^{-2/3} \le t^{-2/3}f(t) \le t^{-2/3}n$
$\int_a^1 mt^{-2/3}\ \mathrm dt \le \int_a^1 t^{-2/3}f(t)\ \mathrm dt \le \int_a^1 t^{-2/3}n\ \mathrm dt$
$3m \le \int_a^1 t^{-2/3}f(t)\ \mathrm dt \le 3n$
and there is where i got stuck

Akiva Weinberger

16:43

like $\sin(1/x)$

DHMO

same place as @GridleyQuayle

Balarka Sen

Show that it's a Cauchy sequence or something, probably. For a given choice of a sequence $a_i$ tending to $0$.

Should be doable.

Akiva Weinberger

Maybe look at the derivative wrt $a$?

I don't know if it works, but it's an easy computation, right?

DHMO

@AkivaWeinberger but it is not continuous at $x=0$

Balarka Sen

@DHMO He's speaking that it's theoretically possible until you prove it.

Akiva Weinberger

16:45

@DHMO It's not defined there, that's my point

DHMO

@BalarkaSen I know

@AkivaWeinberger but f is continuous on [0,1]

Akiva Weinberger

@DHMO But $\int_a^1t^{-2/3}f(t)dt$ isn't defined at $a=0$

So theoretically that function could look like $\sin(1/a)$

DHMO

@AkivaWeinberger i'm saying that the question requires $f$ to be continuous on $[0,1]$

Balarka Sen

I think it's easy enough to show that it's a Cauchy sequence, @Gridley.

It's literally the same argument.

Gridley Quayle

That what is Cauchy? You mean consider a sequence of integrals?

Akiva Weinberger

16:48

@DHMO So?

$f$ isn't the function I'm worried about

DHMO

@AkivaWeinberger so $\sin(1/x)$ doesn't satisfy that requirement

oh

Akiva Weinberger

@DHMO I wasn't comparing $f$ to that function, I was comparing the integral to it

DHMO

So $\lim\limits_{x\to0} \sin(1/x) x^{2/3} = 0$

Astyx

Let $E = C^0([0,1], /Bbb R)$ with the infinite norm $||\cdot||_{\infty}$ and $\phi:\begin{cases}E\to \Bbb R\\ f\mapsto \int_{0}^{1/2}f - \int_{1/2}^1 f\end{cases}$, and finally $H = \ker \phi$. How should I go about computing the distance from $id$ to $H$ ?

Balarka Sen

@Gridly Right.

DHMO

16:50

We can define a function based on $x^{2/3}\sin\left(\dfrac1x\right)$

would it be a counter-example?

no, because $\displaystyle\int_0^1 \sin\left(\dfrac1x\right)\ \mathrm dx = 0.504\cdots$

Gridley Quayle

Thanks, I'll give that a shot. Another quick question, for a space to be Banach, must every cauchy sequence converge uniformly, or just converge?

Akiva Weinberger

This thing (the integral) has a bounded derivative, right?

The derivative being $-a^{2/3}f(a)$.

Mike Miller

just converge. "converge uniformly" doesn't make sense in an arbitrary metric space

DHMO

@AkivaWeinberger bounded derivative implies convergent integral?

Akiva Weinberger

Maybe?

Intuitively it seems true

DHMO

16:53

I do not rely on intuition

Akiva Weinberger

As well you shouldn't

prodprod

isn't it enough to show that the improper integral is absolutely convergent? it should imply the convergence of the improper integral itself

and the argument gridley gave earlier shows that it's absolutely convergent, no?

Balarka Sen

Sure, that's what we are doing.

@prodprod No? It just says the absolute integral is bounded.

prodprod

well ok, technically you still have to say a bit more

namely that the integrals of the absolute values increase as $a$ approaches 0...

Gridley Quayle

17:11

is that roughly correct?

Akiva Weinberger

So, MathCamp is asking us if we have any original math puzzles we can send them for them to put on this year's Qualifying Quiz. So I now might want to solve the meta-problem of coming up with an interesting problem.

This raises the question — what other meta-problems are there?

Astyx

What kind of puzzles ?

Mike Miller

ask for a proof of uniformization

Akiva Weinberger

@Astyx For reference, these are the puzzles from last year: mathcamp.org/2016/qquiz.php

I've solved the 2014 ones

Balarka Sen

@GridleyQuayle I didn't look at that but you seem to have proved that $I_{1/n}$ is Cauchy. What I want to do however is to show is $I_{x_n}$ is Cauchy for any sequence $x_n \to 0$. The thing is you want to show $I_a$, as a function of $a$, is Cauchy-continuous in $[0, 1]$, which might be easier to do. Then continuity is a corollary.

I'll warn that I haven't tried this approach, so no guarantees it'd work.

k, gotta go

Danu

17:43

Hi @MikeMiller

Mike Miller

h

Danu

I'm trying to get better at understanding SES's of vector bundles.

Mike Miller

you put the one thing in then you get rid of it

easy as that

Danu

haha

Sure.

Given a SES $0\to L\to E \to F\to 0$ where $L$ is a line bundle, I'm trying to understand how to get $0 \to L\otimes \bigwedge^{p-1}F\to \bigwedge^p E\to \bigwedge^p F\to 0$

I thought I should do it through understanding transition functions, maybe.

Then I realized I didn't understand transition functions of $\bigwedge^k E$ in terms of those on $E$.

Mike Miller

start topologically. if $E = F \oplus L$, we have $\Lambda^p E = \Lambda^p(F \oplus L) = \oplus_n \Lambda^nF \otimes \Lambda^{p-n} L = \Lambda^p F \oplus \Lambda^{p-1} F \otimes L$

so hey, it works out

that's all i got for ya

Danu

17:50

Hmm...kay

Mike Miller

be my guest on transition functions

Dan B

Okay guys. I'm trying to come up with the best way to get feedback on my Undergrad study activities and mock exam questions. Do you think I could post my workings to SO just to get feedback (and obviously corrections if needed)?

Akiva Weinberger

I dunno, try it, see what happens

Put an "I'm not sure if this belongs here" disclaimer on it

Dan B

I'll give it a go!

Good idea!

Mike Miller

I would vote to close immediately. Try reddit.

Dan B

18:02

Mike! You're a legend, was trying to work out where to put it!

Akiva Weinberger

@MikeMiller My thinking was, even if it got voted to close, people would comment saying where it should get asked instead.

Ahmed Amir

18:48

@Ramanujan I was just curious about the problem you were discussing..

Semiclassical

19:03

@Danu Out of curiousity: Of the complex geometry stuff you're doing (Huybrechts etc) is any of it in the realm of Hodge theory?

Danu

@Semiclassical Hodge theory is essentially a black box in this book.

Brody

19:23

Do one-to-one function and one-one function both mean an injection?

Danu

I hate it when people use that terminology :P

Brody

ha, which one?

Tobias Kildetoft

@Brody both are terrible

just like onto

Astyx

I agree

(not that anyone cares)

Brody

@Tobias I can see how one(-to)-one function is annoyingly close to one-to-one correspondence but I don't find onto too offensive

Tobias Kildetoft

19:32

@Brody Consider the statement "$f$ is a function from $X$ onto $Y$" and note how easy the word is to miss

Brody

@Tobias Ohhh, definitely

Mike Miller

19:46

@Semiclassical note that "Hodge theory" and the Hodge whiten are different

the Hodge theorem*

Tobias Kildetoft

@MikeMiller I read that as the Hodge within

Mike Miller

20:08

definitely much cooler

Semiclassical

20:20

@MikeMiller Which should I be thinking of if I'm seeing stuff about Hodge filtration? (The first?)

Balarka Sen

@Brody Yep.

Mike Miller

@Semiclassical Yes. Danu thought you were talking about the latter.

Forever Mozart

@MikeMiller @BalarkaSen I have a question for you

If $X$ is a continuum say that $x r y$ if $x$ and $y$ are contained in a nowhere dense continuum of $X$.

$r$ is an equivalence relation.

Can you think of a metric continuum with a finite number $n>1$ of equivalence classes?

in $[0,1]$ each eq class is a single pt

but I want finite number

Balarka Sen

Hmm.

Forever Mozart

greater than 1 because $[0,1]^2$ is an example with only 1

it is a tough problem I think

fun to think about

Balarka Sen

20:36

Look at, suppose, the topologist's sine curve $\times$ [0, 1]. How do you plan to "connect" stuff in the accumulation plane of the sine wave to any point on the sine wave with a nowhere dense continuum?

Forever Mozart

you can do it

Balarka Sen

Oh, just join by a topologist's sine curve inside it

Forever Mozart

yeah

Balarka Sen

Very cute problem.

Forever Mozart

yes it may even be unsolved

there is a more difficult variation that I know is unsolved

so I wanted to look at this simpler problem

Balarka Sen

20:39

Neat. You want to study connectedness by joining with nowhere dense subcontinua instead of by paths.

Forever Mozart

yes

Balarka Sen

You can also ask for joining by maps from continua with nowhere dense image in general, like you do for paths, say.

Forever Mozart

in an indecomposable continuum the equivalence classes are just the composants

like buckethandle

so its really only interesting for decomposable

Balarka Sen

ah

Sophie

anyone know how to prove $\displaystyle\sum_{k=0}^n {2k \choose k}{2n-2k \choose n-k}=4^n$?

arctic tern

20:49

more symmetrical: $\sum_{k+r=n}\binom{2k}{k}\binom{2r}{r}=4^n$

Sophie

@arctictern this looks like induction. Thanks

arctic tern

hmm

Forever Mozart

that would be my approach

meow-mix

good afternoon

Sophie

21:03

I have a proof of this, but it is a bit ad hoc. Consider the function $f(x)=\sum_{n=0}^\infty {2n \choose n}x^n=(1-4x)^{-1/2}$ then you get that identity by considering the Cauchy product $f(x)^2$

arctic tern

right

that was the first thing that came to mind, was too lazy to look up generating function f(x) though :P

Fargle

Surely there's some long and complicated combinatorial proof.

shai horowitz

21:38

so a transistor is the fundamental part of your computer that holds a 0 "no current" or a 1 "current" in the cpu and memory of all computers.

my question is does a tertiary transistor

one capable of 3 independent states

must be larger then a binary transistor?

this question stems from the end of moores law and information theory which shows tertiary computing is faster then binary computing

i.e hypothetically in the optimal for both cases does tertiary computing really allow for a greater upper bound in processing power of a cpu of certain size then binary transistors.

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