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July 23[edit]

Advanced Parametric Equation in 3D[edit]

I have also posted my query on some other math oriented boards and a couple science boards in hopes of getting an answer. With that said; Here lie my quandary.
I have been trying to unravel this one for a very long time, ~6 months, and as of yet only have appeared to have come close twice only to find it did not stand up to testing.

This combines both Trigonometric and Astrometric sciences together.

Given an orbital path defined by the minimum constraints such that: M=Semi-Major Axis N=Semi-Minor Axis I=Angle of inclination from the Elliptic plane of the system. L=Angle of Ascending Node (Herein Rendered of non importance to the final Equation)

Under the Condition of I= 90 deg the corresponding surface would be comprised of 2 nested spheres with the outer Sphere having a Radius of M and the Inner Sphere having a Radius of N.
Under the Condition of I<90>0 would be a Sphere Negated in part by a Hyperboloid of one sheet, such that the Sphere has a Radius of M and the focii of Hyperboloid is dependant on I.
Under the Condition of I=0 the structure would degenerate in to a 2D torus such the the outer Circle has Radius M and the Inner Circle is of radius N.

I am hoping that someone will know the parametric equation set for {x,y,z} to create this construct.

I do have a possibly simpler description. Take a Torus M-N of inclination I and Revolve around z axis.

I know the parametric set for a torus of I=0 is: { x=x0+(M+N*cos(u))*cos(v), y=y0+(M+N*cos(u))*sin(v), z=z0+(N*sin(u)) }

I know neither how to incline the torus nor how to perform the second Transform of revolution. Abyssoft 18:14, 23 July 2007 (UTC)[reply]

Since your surfaces all appear to have cylindrical rotational symmetry, it will be much easier to work in cylindrical co-ordinates r, θ, h, and only convert to Cartesian co-ordinates if and when necessary. In cylindruical co-ordinates the equations of the surfaces that you describe in your first and third cases are:

r^{2}+h^{2}=M^{2}{\mbox{ or }}N^{2}

r^{2}+h^{2}=r(M+N)-MN

Because of the cylindrical symmetry, that there is no dependence on θ. I do not understand your description of your second case well enough to determine an equation for this surface. Gandalf61 13:17, 23 July 2007 (UTC)[reply]

What is the origin of the conditions on the desired manifold? Would you be happy with any solution meeting these requirements, or should it in some sense be "natural"? If the latter, one would expect the results for the cases I = 0° and I = 90° to be the limits of the general case 0° < I < 90° as I approaches these limiting values. However, from your description – which is not perfectly clear; what is the meaning of "negated" here? – there appears to be a jump discontinuity there. --Lambiam 15:46, 23 July 2007 (UTC)[reply]

Sorry to be clear as mud, upon further reflection of the problem, the simpler description meet my needs. All the formulae I have already are Parametric, and I am terrible with transforms to be honest. As for Limits on I >= 0° and I <= 90°.
I'll restate the simpler description.

Take a Torus M-N of inclination I and Revolve around z axis.

I know the parametric set for a torus of I=0 is: { x=x0+(M+N*cos(u))*cos(v), y=y0+(M+N*cos(u))*sin(v), z=z0+(N*sin(u)) }

I know neither how to incline the torus nor how to perform the second Transform of revolution.
I wish I could describe it a lot better as I can see it in my minds eye but I'm having a devil of a time getting it from brain to hand(mouth) so to speak. Abyssoft 18:14, 23 July 2007 (UTC)[reply]

If you take a torus and rotate it around another axis than its own axis of rotational symmetry, you actually trace out a complicated solid object in space. Do I understand correctly that you looking for a parametric description of its surface (topological boundary)? --Lambiam 19:31, 23 July 2007 (UTC)[reply]

That is precisely what I'm looking for as it accurately describes boundaries of the orbital field. Modification to M and N based on the Hill Radius results in an accurate description of the boundary of the Gravitation field of Primary influence. in order to help you help me I should Provide a Truer equation set for the Torus involved. Again I unfortunately do not know how to apply the inclination transform so I will make do without that part.

R_Hill=Hill Radius of the body in question.

e of the system only defines the base for M and N.

r=R_Hill/M

{

x=(M+N*cos(u))*(1+r)*cos(v),

y=(M+N*cos(u))*(1+r)*sin(v)*,

z=((N*(1+r))*sin(u))

} This is unfortunately as Far As I have been able to get before my brain melts down. Abyssoft 01:53, 24 July 2007 (UTC)[reply]

Boolean Algebra[edit]

As a pet project I'm trying to optimize the logic for a BCD-to-7-segment-display decoder. I've optimized it pretty far but I'm having trouble with the logic for segment D. So far what I have is (and ABCD represent the bits of the BCD digit from high to low significance):

(/B + (C XOR D)) * (B + C + /D)

The thing that's so tantalizing is that it's basically:

(B OR SOMETHING1) AND (NOTB OR SOMETHING2)

Is there some way to take advantage of this B-NOTB relationship and optimize it further? It feels like if I can rearrange it somehow then I can also take advantage of the Cs and Ds on both sides of the conjunction --⁪froth^t 15:47, 23 July 2007 (UTC)[reply]

For the digit 9, A = D = 1 and B = C = 0, so your formula evaluates then to (/0 + (0 XOR 1)) * (0 + 0 + /1) = 1 * 0 = 0. But shouldn't segment D be on for digit 9? --Lambiam 16:32, 23 July 2007 (UTC)[reply]

There's no tail on the 9 (putting a tail on the 6 and not the 9 just makes things optimize beautifully- you can use a full half of the K-map as a single disjunction for the tail segment on the 6 and you can use an XOR for the 9) . All the other digits work. But any ideas on how to optimize? >_< --⁪froth^t 17:09, 23 July 2007 (UTC)[reply]

All devices I could find in my neighbourhood using 7-segment digits (microwave, video recorder, several alarm clocks) display the digit 9 with a tail, thus:

. I assume then you have a special reason for wanting the tailless

? --Lambiam 19:06, 23 July 2007 (UTC)[reply]

I forgot nines have tails when I was mapping it out. Also like I said it turns out to be better for logic minimization if it doesn't have a tail. --⁪froth^t 19:47, 23 July 2007 (UTC)[reply]

OK through guesswork and a lot of truth tables, I figured out that

(B OR SOMETHING1) AND (NOTB OR SOMETHING2)

is logically equivalent to:

(B AND SOMETHING2) [X]OR (NOTB AND SOMETHING1)

(the reasoning is that if B then SOMETHING2 needs to be true for the expression to be true, and if /B then SOMETHING1 needs to be true for the expression to be true. the or was a half-educated guess) Like some kind of weird de morgan's transformation. It's an XOR because they could never both be true, but it doesn't really matter. Is there a name for this type of rule? And does anyone see any further optimization, or is this a dead end?

(/B * (C + /D)) XOR (B * (C XOR D))

--⁪froth^t 17:42, 23 July 2007 (UTC)[reply]

Maybe you should have a look at our article Boolean algebra (logic), which gives the laws you can use in manipulating and simplifying Boolean expressions. I don't know a name for these specific transformations, but they are readily derived from the Boolean algebra laws. In digital logic, XOR is not a pleasant operator to implement, unlike (N)AND and (N)OR. --Lambiam 19:19, 23 July 2007 (UTC)[reply]

What do you mean not pleasant to implement? As long as you're not actually building the gate, how is it any different than any other logical operation? --⁪froth^t 19:41, 23 July 2007 (UTC)[reply]

I meant indeed building the gate out of transistors; you need many more than for an OR gate. Another issue is that if the two inputs to an XOR gate flip synchronously, but there is a tiny time difference for the changes to arrive at the gate, it can briefly output a spurious value. For this application that it is not an issue, but for some applications it may be. --Lambiam 03:22, 24 July 2007 (UTC)[reply]

Very interesting. This is purely a logic exercise so you're right it wouldn't matter- besides, that's why circuits are clocked --⁪froth^t 05:22, 24 July 2007 (UTC)[reply]

Interesting results from my continuing work.. the equation can be compressed all the way down to C + (B EQ D), however then the 7 has a tail. I for one welcome our new digit :) And astonishingly, directly connecting /D results in near-perfect results as well, but it's still flawed. --⁪froth^t 02:40, 24 July 2007 (UTC)[reply]

Aha! I do believe I have got it!

(/B + (C XOR D)) * (B + C + /D)

EQ

(/B * (C + /D)) XOR (B * (C XOR D))

EQ (drum roll please...)

(D EQ (B XOR C)) + (/B * C * /D)

Doesn't look a lot more minimized than the first expression does it? Oh well, I'm satisfied since I can share B XOR C with the logic for segment G. Thanks for letting me use your math desk for my musings :) --⁪froth^t 02:57, 24 July 2007 (UTC)[reply]

Also in case anyone's curious here are my mostly final results:

A: A + C + (B EQ D)

B: (C EQ D) + /B

C: A + B + /C + D

D: (D EQ (B XOR C)) + (/B * C * /D)

E: /D * (/B + C)

F: (/C + (B * /D)) * (A + B + /D)

G: A + (B * /D) + (B XOR C)

If anyone sees any glaring possibilities for further minimization please point them out but otherwise this is my final solution --⁪froth^t 03:04, 24 July 2007 (UTC)[reply]

Knuth's Pre-Fascicle 0c (Boolean evaluation) for TAOCP volume 4 has the 7-segment display as a worked-out example. Download it fast for the pre-fascicles tend to be taken down the net after some time. – b_jonas 09:34, 24 July 2007 (UTC)[reply]

Grr, postscript. OK I'm not pretending to understand how he reached his results, but they don't seem too impressive. For example, I worked out one chain of logic and here's his expression for segment A:

[(B XOR C) XOR (/A * D)] * [C * (B XOR C XOR D)]'

This is obviously far larger than it needs to be, but his solution is so well-distributed into shared steps that if implemented his would actually have about the same number of gates as mine. Annoyingly he arbitrarily inverts his steps without counting the cost of an inverter, but it wouldn't add a lot if counted. Remarkably (and I'm astonished by the incredible efficiency of his sharing) his entire system adds up to 22 logical operations- count my little plusses and EQs etc above and you get 24.. however I don't just arbitrarily invert without counting an invert, and (B * /D) can be shared between F and G, and (B XOR C) can be shared between D and G, so mine actually uses less logic =D. Unfortunately I think his point was just to demonstrate his logic-sharing process (unlike mine which was for sheer minimization) so it's not a fair contest. Nice find on the link by the way, do you actually read all of that? --⁪froth^t 16:06, 24 July 2007 (UTC)[reply]

The logic described in the fascicle implements the hex digits A-F, not just 0-9, which is why the logic implemented may seem somewhat redundant.

Well it seems large because of the extensive term-sharing, but if it indeed is a full solution for 0-f then that beats the pants off mine :( --⁪froth^t 05:23, 25 July 2007 (UTC)[reply]

Just read the whole thing. There is a solution for just 0-9 too later. He inverts terms because his model takes all 16 two-input one-output binary operations of the same cost, so negated inputs just means a different basic operation. (My opinion is that this is realistic, because if you have not only AND, OR, XOR but also a fourth one like implication as basic operations, then you basically have all of those 16 operations only sometimes you get a negated result but you can keep track of that and only ever have to negate only the final results.) – b_jonas 10:28, 25 July 2007 (UTC)[reply]

No you can't just negate the final results..what about in the case of ( A OR B ) and both A and B are actually inverted? A' OR B' isn't the same as (A OR B)' it's the same as (A AND B)'. This is additional logic and isn't simply a boolean flip to be counted up and turned into either a single inverter or no inverter at the end of the chain. --⁪froth^t 00:51, 29 July 2007 (UTC)[reply]

Spaces of p- and q-summable sequences[edit]

Hi all, yet another real analysis question vexes me. Perhaps it shall vex thee, and ye can help me. =)

Suppose that $1\leq p<q<\infty$ and suppose we have a sequence x in $\ell ^{p}\cap \ell ^{q}$ . Is there an inequality between the norms $\|x\|_{p}$ and $\|x\|_{q}$ valid for all such x? Are the two norms equivalent?

Now, I know what everything here means, and the first part of the question is to do the same thing for $L^{p}[0,1]$ and $L^{q}[0,1]$ , which I can (and have recently) done. However, this is just really throwing me.

Through various trials and errors, I have come to the conclusion that it's probably the case that $\|x\|_{q}\leq \|x\|_{p}$ , but I don't know how to show this or how to prove or disprove the equivalency of the norms. Any aid would be appreciated! –King Bee ^{(τ • γ)} 20:32, 23 July 2007 (UTC)[reply]

The norms are clearly not equivalent, as there are sequences for which one is finite and the other is not (so even if you restrict to

\ell ^{p}\cap \ell ^{q}

, the ratio can be aribtrarily large). As for

\|x\|_{q}\leq \|x\|_{p}

, I'm pretty sure there's some well-known inequality which gives this instantly, but I don't remember which. Jensen's inequality probably solves this easily, though. A different approach is to first prove

{\sqrt[{p}]{a^{p}+b^{p}}}\geq {\sqrt[{q}]{a^{q}+b^{q}}}

(for which Binomial expansions are handy), generalize to any number of summands, and pass to the limit. -- Meni Rosenfeld (talk) 21:07, 23 July 2007 (UTC)[reply]

Another solution: First normalize the vector so that its greatest coordinate is 1. Then, not only does the q-norm raise each coordinate to a higher power, thus decreasing it, at the end it takes a higher root, decreasing the result further. -- Meni Rosenfeld (talk) 21:13, 23 July 2007 (UTC)[reply]

From the inequality it follows, of course, that

\ell ^{p}\subset \ell ^{q}

, that

\ell ^{p}\cap \ell ^{q}=\ell ^{p}

and that it is the p-norm that can be arbitrarily greater than the q-norm. -- Meni Rosenfeld (talk) 21:18, 23 July 2007 (UTC)[reply]

Technically, we know that all norms on a finite-dimensional vector space are equivalent, but we also know that the sequence (1,1,1,…) has a p-norm for every p except p = 1. We also know that a sequence space is a special case of a function space, using the Lebesgue integral with counting measure over N. So that's two lines of attack.

But more important, especially on test day, is the part about "this is just really throwing me". You would not have come this far without intellect and knowledge, and you have studied diligently. Go in knowing you are capable of passing, and don't need to be perfect to do so. A test is scribbles on paper, nothing more. Try swimming, music, yoga, whatever helps you relax and calm your mind. Prepare as well as you can, get a good night's sleep, then let it all go and just do the best you can. --KSmrq^T 22:15, 23 July 2007 (UTC)[reply]

Do we really know that the sequence (1,1,1,…) has a p-norm for every p except p = 1? Perhaps it should be (1/1,1/2,1/3,…)? -- Meni Rosenfeld (talk) 22:18, 23 July 2007 (UTC)[reply]

Oops, sorry; yes, of course. The ones sequence is obviously unbounded for every p, the harmonic one is what I meant to say. --KSmrq^T 01:58, 24 July 2007 (UTC)[reply]

This is a classical exercise, and something that is neat to remember. The

L^{p}

spaces are nested on spaces of measure 1 - and probably on all (totally) finite measure spaces - (apply Hölder's inequality); with the counting measure, the inequality on the norms (and thus inclusion of the spaces) is reversed. To see this, use the fact that we have

\sup |f|^{p}\leq ||f||_{p}^{p}

, and write

|f|^{q}

as

|f|^{p}|f|^{q-p}

. I guess I'm not adding anything new here, really. These inclusions are cute, but their usefulness is quite limited since the are (of course!) not available on the line; which is a drag, really (imagine what Fourier analysis would look like!). Phils 15:27, 24 July 2007 (UTC)[reply]

KSmrq - Thanks for the advice. It's often hard to keep these things in perspective sometimes. I do get to the gym every morning at around 7 to run around and get some exercise, which is helpful for the mind in the long run. I'll try to remember this and stay calm during the actual exam. =)
Meni - Your "alternate solution" doesn't really work (as far as I can tell). After dividing x by $\|x\|_{\infty }$ , I get this extra factor of $\|x\|_{\infty }$ floating around. I might be misunderstanding you, so here's what I have:

Pick $x\in \ell ^{p}\cap \ell ^{q}$ . Then $x\in \ell ^{\infty }$ , so $\|x\|_{\infty }<\infty$ . Assume $\|x\|_{\infty }\geq 1$ , so $\sum |x_{i}|^{p}\geq 1$ . As p < q, we have

\sum \left({\dfrac {|x_{i}|}{\|x\|_{\infty }}}\right)^{q}\leq \sum \left({\dfrac {|x_{i}|}{\|x\|_{\infty }}}\right)^{p}

, which gives

\|x\|_{q}\leq \left(\|x\|_{\infty }^{q-p}\sum |x_{i}|^{p}\right)^{1/q}\leq \|x\|_{\infty }^{1-p/q}\|x\|_{p}^{p/q}\leq \|x\|_{\infty }^{1-p/q}\|x\|_{p}

And I'm stuck.

Also, this "taking a higher root" deal only works if the sum $\sum |x_{i}|^{p}$ is greater than 1 (which it very well may not be). Suggestions? Am I missing something so obvious here? –King Bee ^{(τ • γ)} 15:19, 24 July 2007 (UTC)[reply]

You basically got it, just notice that

||x||_{\infty }^{p}\leq ||x||_{p}^{p}

. Now from what you have, and that inequality, we get

||x||_{q}^{q}\leq ||x||_{q}^{q-p}||x||_{p}^{p}=||x||_{p}^{q}

. Note that there are no worries about taking q-th roots here. The point is that the q-th root function is increasing so it preserves inequalities. That it has the behavior we know on

(0,1)

is irrelevant. Phils 15:43, 24 July 2007 (UTC)[reply]

[edit conflict]Yes, you have missed several very obvious things. I will denote

M=\|x\|_{\infty }

and

y={\frac {x}{M}}

, so that

\|y\|_{\infty }=1

.

Since $\|y\|_{\infty }$ = 1, there is some $|y_{i}|=1$ , so $\sum |y_{i}|^{p}$ has 1 as one of its summands and all others are non-negative. Thus $\sum |y_{i}|^{p}\geq 1$ (where the inequality is clearly strict if there is any other nonzero coordinate). Thus $\|y\|_{q}\leq \|y\|_{p}$ (where y has ∞-norm 1).
$\|\cdot \|_{r}$ is a norm. So multiplying a vector by a positive scalar multiplies the norm by that scalar. Therefore $\|x\|_{q}=M\|y\|_{q}\leq M\|y\|_{p}=\|x\|_{p}$ . It's really that simple. I suppose your calculation above failed because you have used inequalities which are too weak.

-- Meni Rosenfeld (talk) 15:53, 24 July 2007 (UTC)[reply]

Thanks for spelling it out. Gah. I thought it was going to be a little more technical. Please excuse my dullardness. –King Bee ^{(τ • γ)} 16:04, 24 July 2007 (UTC)[reply]