Mathematics - 2017-01-23 (page 3 of 4)

8:00 PM

I told myself I was going to focus on numerical analysis and not get distracted by other math tonight. I already failed twice :P

Ted Shifrin

You're extremely good at such failures, @Alessandro. It's one of your strengths, I would say :P

I'm currently distracted by a differential geometry question that Balarka could have asked but didn't.

Alessandro Codenotti

Indeed, I must agree

that must be a very interesting question to be distracting before being asked

Ted Shifrin

But I think it's one of the measures of what kind of mathematician you are, @Alessandro.

@AlessandroCodenotti I figured I should ask Balarka the question he should have asked, but I'm curious to know the answer :P

I'm pretty sure I know the answer is no, but I want an example.

Alessandro Codenotti

@TedShifrin I'm not sure how to interpret that

ah, I see, he'll probably be around soon

Ted Shifrin

(It was intended as a compliment.)

Yes, it's getting to be his bedtime, so he probably will show up.

user379685

8:05 PM

@TedShifrin I'm lost :c

Ted Shifrin

Heya @Daminark

Daminark

Hey @Ted, how's it going?

Ted Shifrin

Good linear algebra question for you :)

Daminark

Do tell!

Ted Shifrin

If the characteristic polynomial of $A$ is $-t^3-2t^2+t+2$, what is $\det(A^3+A^2-A-I)$?

Alessandro Codenotti

8:06 PM

$-I$ at the end

user379685

@TedShifrin how can i find pb(1)?

Hippalectryon

@TedShifrin \o

Ted Shifrin

@user379685: $p_B(t) = -(t-\mu_1)(t-\mu_2)(t-\mu_3)$, where the $\mu_i$ are the eigenvalues of $B$, right?

Salut, @Hippa!

Alessandro Codenotti

(well, the approach should be the same with $+I$)

Ted Shifrin

Thanks, @Alessandro.

Hippa probably knows a good way to do this problem.

user379685

8:08 PM

@TedShifrin yes

Hippalectryon

@TedShifrin Haven't done any maths in some months now >.> what's the problem ? this one ?

2 mins ago, by Ted Shifrin

If the characteristic polynomial of $A$ is $-t^3-2t^2+t+2$, what is $\det(A^3+A^2-A-I)$?

Ted Shifrin

$p_B(1)=-(1-\sum \mu_i + \sum_{i<j} \mu_i\mu_j - \mu_1\mu_2\mu_3)$. @user379

user379685

uhhhh

Ted Shifrin

So can you figure out those expressions in $\mu_i$ from knowing stuff about $A$? @user379685

Yes, @Hippa.

I will say that usually these problems are made up to be easy if you do something obvious, but I don't see that here (and neither did @Alessandro).

user379685

@TedShifrin how did you get that formula?

Ted Shifrin

8:10 PM

I multiplied out the factored formula I had before.

Which is what you should do with a general characteristic polynomial to realize that the coefficients are expressed in terms of the eigenvalues of the matrix.

Alessandro Codenotti

yep that seems uglier than the usual problems but I might be missing something of course

Hippalectryon

Uh >.> I probably would have been able to do something a few months back, but it's really fuzzy in my head rn :( i'd need to go back to my class notes a bit

Daminark

So $1$ and $-1$ are eigenvalues

Alessandro Codenotti

Are there nice interpretations for the coefficents of the terms of degree $2$ of higher of the characteristic polynomials as there are for the degree $1$ and $0$ terms?

Ted Shifrin

Oh there's another typo. The characteristic polynomial is $-t^3-3t^2+t+2$, which has no integer root.

Daminark

8:13 PM

And because the polynomial is cubic, the third root is real. It feels like it's going to be different, so I'm gonna try to find it, then I guess one (unelegant) way would be to manually calculate the determinant of that expression using a diagonal form of $A$

Oh

Ted Shifrin

@Alessandro: You don't mean $1$ in general.

You mean $n-1$. That's (up to sign) the trace. Yes, the coefficients are all, up to sign, the symmetric polynomials in the eigenvalues :)

Super super important, even for topology and geometry.

Sorry, @Daminark. The folks should have caught that.

Daminark

It's alright, haha

Ted Shifrin

@user379685: First question I asked you: Do you know the eigenvalues of $A^3+A^2-A$ in terms of the eigenvalues of $A$?

user379685

wait

Ted Shifrin

I've been prone to errors today, it seems, @Daminark. Time for me to retire. Oh.

Alessandro Codenotti

8:15 PM

@TedShifrin I see, interesting

user379685

@TedShifrin No i do not :(

Ted Shifrin

Hint: If $Av=\lambda v$, what is $(A^3+A^2-A)v$?

user379685

it's λv(A^2+A-A) ?

Ted Shifrin

(@user379685: What sort of course are you in?)

user379685

Linear algebra

Ted Shifrin

8:19 PM

Agh. That is not making any sense at all.

What kind of linear algebra class?

What you wrote makes me think this problem is too hard.

user379685

I don't know what kinds are there

It was on a exam from a previous year

and it shouldn't be hard

Ted Shifrin

This is a pretty sophisticated problem.

Go ask your professor how to do it.

Daminark

Ah, so it's $(\lambda^3 + \lambda^2 - \lambda)v$

And we know that $\det(A) = 2$

Ted Shifrin

Right. So the eigenvalues of $q(A)$ are $q(\lambda)$ for any polynomial $q$ and eigenvalue $\lambda$ of $A$.

But be careful, @Daminark.

Daminark

So we try to find an expression for the product of the eigenvalues of $A^3 + A^2 - A - I$ in terms of those of $A$?

Also using the other terms of the characteristic polynomial

Ted Shifrin

8:22 PM

I'm guessing that there's a typo in the problem, @user379685. It's too hard for an elementary linear algebra course.

Daminark

Oh, am I nearing a trap?

Ted Shifrin

Not quite, @Daminark.

Well, yes, you want the product of those eigenvalues.

user379685

home.agh.edu.pl/~gora/algebra/Egzaminy.pdf page 6 question 4

Ted Shifrin

I thought you were going to use $\det(A+B)=\det A+\det B$, @Daminark. But you're too smart for that.

Daminark

Ah, haha, yeah, I've made that mistake enough to avoid it

Also

user379685

8:25 PM

@TedShifrin could you quickly summarize your solution, i'm a bit short on time and can't really ask my professor

Daminark

If you use Cayley-Hamilton, you get that $A^3 + A^2 - A - I = I - 2A^2$ if I didn't mess up

That might be easier to work with

Ted Shifrin

From what you wrote down when I asked you about $(A^3+A^2-A)v$, @user379685, I think you have some basic things to learn.

Yes, @Daminark. That'll make the algebra easier.

So you need the product of the eigenvalues of that matrix.

user379685

@TedShifrin like what?

Ted Shifrin

Like how to correctly compute that quantity, @user379685. What you wrote was nonsense.

Balarka Sen

I think I am going to get a fever again. Darn it.

Ted Shifrin

8:28 PM

STOP that, @Balarka.

Did you read above that I'm stuck on a question you should have asked me? :)

user379685

A^3v + A^2v - Av = A^2*Av + AAv -Av = A^2*λv+ Aλv - λv ?

Ted Shifrin

Keep going, @user379685.

Balarka Sen

@TedShifrin Currently reading this. Why's that true?

My intuition is failing.

user379685

=(A^2 + A - I)λv=?

Ted Shifrin

I want no $A$'s left in the answer, @user379685.

It's one of the Codazzi equations, @Balarka.

But the question is: Must the surface be a surface of revolution?

Now that I look at those exams, @user379685, I understand you're in Poland (I think) and such courses are far more sophisticated there than in the US.

user379685

8:32 PM

yeah i am from poland

Maks

@TedShifrin ! Hi

Ted Shifrin

Yeah, they expect serious math students there ... not like most of the places in the US.

hi @Maks

Maks

Could you give me a hand here ?

Ted Shifrin

So I no longer think it is a typo, @user379685. You need to know more than you seem to know.

I'm leaving in a minute, @Maks. What's up?

Maks

I got an $ f(x) = x + cos(x) $, and I have to find its primitive function on (0,4)

user379685

8:34 PM

@TedShifrin can't you just tell me the solution and i'll work on what i'm missing

Balarka Sen

@TedShifrin Because $d\omega_{23} = k_1 d\omega_2 = 0$, so if we feed $e_2$ to it we get $k_1 = 0$ in the $e_2$-direction. Is that what you are saying?

Maks

But, primitive = integral, and I cant calculate the integral on one point, well I mean, I can, but its the value of the function on that point, no need to integrate

And secondly, the function doesnt have the point (0,4)

So what does it mean ??

Ted Shifrin

Interval $[0,4]$, @Maks, not point.

Maks

@TedShifrin it says point..

Ted Shifrin

No, @user379685. It's too involved.

I don't have time to type out a 15-minute answer.

Maks

8:36 PM

OH ! I get it, its the point (0,4), he wants my constant to be 4

So the primitive valuated on 0 is 4

Ted Shifrin

Oh, your English is wrong, @Maks. When you say primitive function on (0,4) ... that is not what that means.

A primitive whose graph goes through (0,4) is different.

user379685

@TedShifrin well, i don't even think it requires so much work, my professor said all of these questions should have an easy and relativly quick solution

Maks

Oh, ok, thank you !

Ted Shifrin

Well, then go ask your professor, because we don't see it.

But in the meantime you had better know INSTANTANEOUSLY what $(A^3+A^2-A)v$ simplifies to if $Av=\lambda v$.

@Daminark, @Alessandro: Did you figure it out?

Daminark

Modulo computation errors I get that the determinant is $2(\lambda_1^2\lambda_2^2 + \lambda_1^2\lambda_3^2 + \lambda_2^2\lambda_3^2 - \lambda_1^2 - \lambda_2^2 - \lambda_3^2) - 7$

Sorry typo

Ted Shifrin

8:39 PM

YOu're missing some $2$'s ...

I don't get the 7. There should be -31 perhaps?

Balarka Sen

@TedShifrin Re your question. If $k_1$ is everywhere constant then it's (a subset of) a tube, by a theorem we discussed a long time ago. So it's indeed natural to expect it'd be a surface of revolution if $k_1$ is constant along the $e_2$-direction (given a principal frame).

Fun.

Daminark

Oh wait no I screwed up the computation big time

I dropped a lot of 2s

That's where the 7 came in as well

Revised answer coming in a bit

Ted Shifrin

@Balarka: $k_1$ is only constant in the $e_2$ direction. It can definitely vary along the $e_1$ curves!!!

You're confused re that previous tube question.

Balarka Sen

I know

No, I am just saying if it's everywhere constant then it's a tube.

Ted Shifrin

In our notation, we'd need $k_2$ constant on the $e_2$ curves, and I cannot deduce that.

Balarka Sen

8:41 PM

I didn't claim your thing is a tube too

Ted Shifrin

I'm trying to show it's a general surface of revolution, but I think that too is false.

So you need to finish, @Daminark :P

Liad

Hi, i am proving that the variance of cauchy distribution is not defined.
i have proved that the expectation of it is not defined , is it correct to conclude that the variance is not defined ?

Daminark

OK so it's actually $4\lambda_1^2\lambda_2^2 + 4\lambda_2^2\lambda_3^2 + 4\lambda_1^2\lambda_3^2 - 2\lambda_1^2 - 2\lambda_2^2 - 2\lambda_3^2 - 31$

Balarka Sen

I'd be surprised if there's a counterexample, but my intuition about these sort of things is terribad

Nice question however

Ted Shifrin

Keeping things grouped/factored will help, @Daminark. Now finish :P

@Balarka: By the Fundamental Theorem of Surface Theory, if we can write down first and second fundamental form that satisfy Gauss, Codazzi, there exists (at least locally) a surface.

Maks

8:43 PM

And one last question @TedShifrin

Alessandro Codenotti

@TedShifrin I'm not really thinking about it to be honest

Mary Star

Suppose that $C$ is an algebraic closure of $F$, $f\in F[x]$ irreducible and $a,b\in C$ roots of $f$.
We have the field extension $F\leq F[x]/\langle f\rangle$. That means that $F\rightarrow F[x]/\langle f\rangle$ is an homomoprhism, right?
Does it hold that $C$ is also an algebraic closure of $F[x]/\langle f\rangle$ ?

Ted Shifrin

Good, @Alessandro. You have numerical analysis.

Liad

@TedShifrin maybe you can help me ? :-)

Alessandro Codenotti

Better not tell @Ted I'm thinking about differential topology in the other chatroom

Maks

8:44 PM

$ \int a^x = a^x log(a) $
or
$ \int a^x = \dfrac {a^x} {log(a)} $

Balarka Sen

He's not doing numerical analysis though. He's procrastinating on tangent bundles

Ted Shifrin

smacks @Alessandro

Try differentiating your answer and see if you're right, Maks.

Maks

@TedShifrin oh, first one is differentiation second one integration

Ted Shifrin

Right.

Maks

You called it differentiate or derivative ?

call*

Ted Shifrin

8:47 PM

The derivative is the result of differentiating.

Maks

Oh, ok

Liad

i have showed $ E[X]$ does not exists ("infinity" - "infinity") , can i conclude that $Var(X) $ does not exists too?

Secret

(Old news) Meanwhile I have an opposite problem: I rely too much on the maths to derive the physics and thus lost track of the phsical reality

Ted Shifrin

I don't think so @Liad.

Liad

Why not?

Astyx

8:50 PM

Hi chat

Liad

$Var(X) = E[X \ ^ 2 ] - E[X ] \ ^ 2 $

Ted Shifrin

@Daminark: My final answer is ... $-1$ ... which makes me think there's a sneaky way to get this faster. But I have no idea what it is.

Balarka Sen

That's not quite how variance is defined is it

Liad

if $E[X] $ does not exists, then $ E[X] \ ^ 2$ does not exists either

Ted Shifrin

@Liad, that assumes things make sense. But that does not imply $E[X^2]$ doesn't exist.

I need to go eat lunch. Bye all.

Secret

8:52 PM

(Slightly old news) q analogue codes Hmm... it seems most maths structure tend to end up in applications involving computer sciences

Astyx

$E[(X-E(X))^2] = E[X^2 - 2XE[X] +E[X]^2] = E[X^2] -E[X]^2$

Bye ted

Daminark

So I'm gonna need to go to Classics soon but my suspicion wrt the computation is that it'll involve some schemery with the $(x+y)^2 - 2xy = x^2 + y^2$

Liad

if it is, then $E[x] \ ^ 2 = Var(x) - E[X \ ^ 2 ] $ exists!

Balarka Sen

I guess it doesn't matter. Variance is the 2nd moment about the expected value, which does not make sense if the expected value doesn't.

Astyx

What was the question ?

Liad

8:53 PM

you asking me?

Daminark

But I'm not yet sure how to pull it off. I'll get back to it eventually. See you!

Astyx

For instance, yeah

Liad

i proved that $E[X] $ does not exists

i want to conclude that $Var(X)$ does not exists too

Owatch

Hello.

Astyx

I think you can

And I agree with what Balarka said

Liad

8:56 PM

i proved it as follows :
if Var exists, $E[X ] \ ^ 2 = E[ X \ ^ 2 ] -Var(X)$, so if Var exists, then $E[X \ ^ 2] $ doesnt, and so we get a contradiction. @Astyx what do you think?

something feel wrong to me here..

Astyx

I think the result is even stronger than this

Liad

What do you mean?

Astyx

If $L^p$ denotes the space of variables than have a p-th moment, then for all $p$ we have $L^{p+1} \ subset L^p$

What kind of variables are you dealing with ? @Liad

Liad

X is cauchy distribution

i showed $E[X] $ is of the form $ \infty - \infty $ so it does not exists, i thought there is a shurtcut for showing the variance does not exists either

Astyx

So continuous ?

Liad

9:02 PM

yes

Secret

Some commutative ring generalisation. My poor background on topology however means part of the abstract is still incomprehensible to me

Astyx

@Liad I guess you could see that $x^p \le 1+ x^{p+1}$

For positive $x$ of course

Mary Star

When C is an algebraic closure of F and a is an element of C, is C also an algebraic closure of F[a] ?

Secret

9:18 PM

Trajectories escaping to infinity if $f$ is transcendental and meromorphic

Ted Shifrin

@Liad: Just compute the variance using the definition (with the integral).

It clearly is undefined, because you have $$\int_{-\infty}^\infty \frac{y^2}{1+y^2}\,dy.$$

Your argument about expectation is not exactly meaningful.

You need to explain why the improper integral $$\int_{-\infty}^{\infty} \frac y{1+y^2}\,dy$$ does not exist. (Its Cauchy principal value is $0$, however.)

Adeek

sorry @TedShifrin I had to go to attend my reading class.

I give lecture to my professor every week.

@TedShifrin I solved the problem while presenting him the lecture in quadratic forms. I solved it in my head.

@TedShifrin Here is my argument

Ted Shifrin

I don't know what you're sorry about. You mean the homotopy between the identity and the antipodal map?

Adeek

yeah

Semiclassical

That does suggest something an interesting question: if a PDF has undefined $n$th moment, can any of the higher moments past that be well-defined?

Adeek

9:26 PM

Suppose n is odd then show that $\phi : S^n \rightarrow S^n$ antipodal is null-homotopic. This will be null-homotopic if we can construct a continous extension $\psi : B^{n + 1} \rightarrow S^n$. Construct such extension as $x \mapsto \phi(\frac{x}{||x||})$

Ted Shifrin

It seems to follow from comparison if you use Lebesgue integrals (taking absolute values).

Adeek

Then we are done

Ted Shifrin

Whoa, Karim. You're trying to prove the wrong thing.

The antipodal map is never nullhomotopic.

Adeek

oh

Semiclassical

Mmkay

Adeek

9:27 PM

It is homotopic to the identity :S :S

I see

I thought something is wrong...

no spoilers I will figure it out.

Ted Shifrin

What you wrote down clearly makes no sense on the ball. You need to pay attention to crap you write!

Adeek

yeah

Owatch

Really beginning to despise signals.

Ted Shifrin

Turn signals?

Owatch

No. Just sinusoids.

Ted Shifrin

9:35 PM

oh

Owatch

Have to get the components for a couple just from a diagram of the wave. Really frustrating for me when the waves are products of a couple different sine waves.

Just need practice..

Ted Shifrin

You know the $\sin(A)\sin(B) = -\cos((A+B)/2)+\cos((A-B)/2)$ type formulas, @Owatch?

Adeek

I think I have the idea geometrically. I will try to turn my intuition into math. I think I want a homotopy from identity to the antipodal map. I think the way I want this to work is to move from a point anti clockwise into its antipodal point using the non-vanishing vector field somehow so we are moving along a arc. @TedShifrin

Ted Shifrin

OK Karim. Now do it :P

Owatch

Yes I am aware of those formulas.

Ted Shifrin

9:40 PM

So that's how you do a product of two sine waves, @Owatch.

Owatch

But I'm not drawing sinusoids using a product given to me.

I'm doing it the other way around.

Ted Shifrin

Oh, you said products.

Balarka Sen

@TedShifrin So what do you recommend me to do next? I clearly haven't developed the intuition for moving frames I want yet

Ted Shifrin

Have you done all the exercises, @Balarka? I can give you more from my grad course, too, of course.

Balarka Sen

Any exercises you recommend for me?

Owatch

9:40 PM

Yeah, the waves I'm trying to get the components of are products of simpler waves (I'm only given the plot though).

Balarka Sen

Nah, haven't done much

Ted Shifrin

I mentioned the last one as particularly fascinating (Bäcklund's Theorem for producing a surface of constant negative curvature from another one)

Balarka Sen

Ah, ok.

Ted Shifrin

But there are several good exercises, and a few boring ones just to get used to stuff.

Kasmir Khaan

hi chat , when they give me a cylinder or a parapoloid eg. x^2+y^2 =2 , 0<z<3 , and z= x^2+y^2 , 0<z<1 , should I assume that the surface is open when I use the divergence theorem ?

Ted Shifrin

9:41 PM

I actually am curious about a non surface of revolution along the lines we were discussing.

@Kasmir: You keep writing < but it should always be $\le$.

But you cannot apply the divergence theorem unless the surface has no boundary — i.e., encloses a finite volume.

Kasmir Khaan

Can you tell me how you type in latex form that fast?

Ted Shifrin

I have been typesetting papers and books for years.

Kasmir Khaan

and when i count the flux , should i then delete the flux from the surface that i added ?

Ted Shifrin

@Balarka: A lot of those "redo" exercises are excellent for you to do, too.

Correct, @Kasmir.

Balarka Sen

@TedShifrin I don't have much geometric intuition about the exterior derivative of the forms coming out of the moving frames. I can plug the formulas but I feel like I am missing something.

Ted Shifrin

9:44 PM

@Balarka: If you want a cool challenge, prove Clairaut's relation for geodesics on a surface of revolution using moving frames. It takes thinking about actual triangles :P

Kasmir Khaan

the normal have to be the same eg. the out nomral in the case of paraboloid opning upward and a plane z=1 , the normal for the circle at z=1 is ( 0,0,1)

Balarka Sen

Hmm, thanks. I'll bookmark this.

Ted Shifrin

Well, in fairness, what intuition do you have for $P_y-Q_x$?

Kasmir Khaan

but i take negative of that when am done correct?

Ted Shifrin

@Kasmir: All the normals have to be compatibly oriented. It depends on which way the normal to your paraboloid is pointing.

Semiclassical

9:46 PM

Hi chat.

Kasmir Khaan

okay thnaks ! :)

Ted Shifrin

@Balarka: Ultimately, all intuition for the exterior derivative comes from thinking about Stokes's Theorem "infinitesimally."

Kasmir Khaan

Anyone know a good lectures online about complex analysis ?

._.

Ted Shifrin

Hubbard and Hubbard (which I dislike) even defines the exterior derivative that way — or they did in the early editions. I thought that was cruel and unusual for a freshman taking the course.

Balarka Sen

You have a sign error :) I guess in general I think of it as curl of the vector field $(P, Q)$ the form is dual to.

@TedShifrin This is a fair point.

Ted Shifrin

9:47 PM

So you're actually thinking about curl physically? Good for you :P

Semiclassical

$z$-component of $\nabla\times \mathbf{F}$.

Ted Shifrin

Yeah yeah

Astyx

Hi @Semi

Semiclassical

I tend to default to vector calculus, for better or worse.

Balarka Sen

Yeah, I think of it as a measure of how a thing rotates under the action of the vector field.

Ted Shifrin

9:49 PM

Good luck in 4-space :P

curl doesn't make sense

Balarka Sen

Yup, screwed for 2-forms.

Semiclassical

Nope.

Astyx

Is there no equivalent of curl in 4-space ?

Ted Shifrin

@Balarka: The nice form of Codazzi with moving frames reduces to something like that Lemma 3.3 (which you tried to ignore) in section 2.3, which gave a reasonable setting for the horrid Codazzi equations. Ultimately, you want to learn Frobenius and relate d to integrability questions for distributions.

No, @Astyx. There's a 6-component 2-vector.

Nothing resembling a vector field.

Astyx

Oh right

Semiclassical

9:51 PM

Both $dx_1\wedge dx_2$ and $dx_3\wedge dx_4$ are 2-forms.

user379685

@TedShifrin A^3+A^2-A-I=(A+I)^2(A-I)
det(A^3+A^2-A-I)=det^2(A+I)det(A-I)
A+I=B
(A+I)*v=λb*v
λb=λ+1
det(A+I)=detB=λb1*λb2*λb3=(λa1+1)(λa2+1)(λa3+1)=1+(λa1+λa2+λa3)+
+(λa1λa2+λa1λa2+λa2λa3)+λa1λa2λa3 Vieta's formulas and done

Ted Shifrin

Yes, Vietas formulas is what I was using. I didn't think to factor $A^3+A^2-A-I$, however. That was the sneaky thing I missed.

user379685

god i lost 1h of my life :c

Ted Shifrin

Wait. Your factoring is wrong.

Semiclassical

Though, that bothers me a little. One can do electrogmagnetism in either 3+1 spacetime or just 3-space; one does curl in the latter, but it should also show up (in a presumably disguised way) in the former.

Ted Shifrin

9:53 PM

It's (A^2-I)(A+I) = (A+I)^2 (A-I). OK.

You had it right. I just didn't simplify it the same way.

I still say you need to know what I said you needed to know, though.

user379685

oh yeah i forgot about that, you're right xd

Semiclassical

But then, electric and magnetic fields look a bit different in 3+1D than in 3D. (You can write both as 3-vectors in the latter case, but you can't write either as 4-vectors.)

Ted Shifrin

Good luck on your exam. Thanks for showing me the trick :)

user379685

@TedShifrin Thanks, no problem.

Semiclassical

(or at any rate one shouldn't, since they transform as the Faraday tensor.)

Ted Shifrin

9:55 PM

@Semiclassic: Probably best to do E&M with forms (I even put Maxwell's equations in my book) :)

Semiclassical

In 3D or 3+1D?

Ted Shifrin