text:
Changing the heading style of references section

\documentclass[twoside, a4paper, 11pt]{article}
\usepackage{natbib}

\begin{document}

\bibliographystyle{apa-good}
\bibliography{bibliography}
\end{document}

How can I make the section header for my references list bolded, underlined, all capitalized, 11 pt and with C6. in front of it? I.e., currently, the section header is: References however I would like it to be:

EDIT:

\documentclass[twoside, a4paper, 9pt]{article}
\usepackage{natbib}

\renewcommand{\refname}{C6.~REFERENCES}
\makeatletter
\renewcommand\bibsection{%
\section*{\underline{\refname}\@mkboth{\MakeUppercase{\refname}}{\MakeUppercase{\refname}}}%
}%
\makeatother

\begin{document}

\underline{\textbf{C1. PROJECT TITLE}}

blkah blah

\underline{\textbf{C2. AIMS AND BACKGROUND}}

blah blah blah

\bibliographystyle{apa-good}
\bibliography{bibliography}
\end{document}

The output is:

The size of references doesn't change when I change the font size to 9pt. I know I should use \section for section titles but there is a reason why I am manually creating section titles. So how can I get the references heading size to be the same as my manually made ones?

• Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. Dec 8 '13 at 14:50
• What are C and 6 supposed to represent? Are they part of some numbering system, or should C6 simply be hardcoded in the heading? Dec 8 '13 at 14:57

1 Answer

If you are sure that you are using the article document class, just adding the following lines in your preamble should work:

\renewcommand{\refname}{C6.~REFERENCES}
\makeatletter
\renewcommand\bibsection{%
\section*{{\normalsize\underline{\refname}}\@mkboth{\MakeUppercase{\refname}}{\MakeUppercase{\refname}}}%
}%
\makeatother

MWE

\documentclass[twoside, a4paper, 11pt]{article}

\begin{filecontents*}{bibliography.bib}
@article{reference,
author = {F Author and S Author},
journal = {Journal},
title = {Article Title},
year = {2013},
month = {10},
}
\end{filecontents*}

\usepackage{natbib}

\renewcommand{\refname}{C6.~REFERENCES}
\makeatletter
\renewcommand\bibsection{%
\section*{{\normalsize\underline{\refname}}\@mkboth{\MakeUppercase{\refname}}{\MakeUppercase{\refname}}}%
}%
\makeatother

\begin{document}

\nocite{*}

\bibliographystyle{apa-good}
\bibliography{bibliography}

\end{document}

Output:

If you also want to remove the spacing add \vspace*{-.8\baselineskip} in the above definition, so to have

\renewcommand{\refname}{C6.~REFERENCES}
\makeatletter
\renewcommand\bibsection{%
\section*{{\normalsize\underline{\refname}}\@mkboth{\MakeUppercase{\refname}}{\MakeUppercase{\refname}}}\vspace*{-.8\baselineskip}%
}%
\makeatother

Output:

• Thanks karlkoeller, I wanted C6 hardcoded. Also is there a way I can control the size of the title header? E.g, what if I wanted it in 20pt? Dec 8 '13 at 15:08
• @user42388 Try adding \Huge (or \huge or \LARGE) before \underline Dec 8 '13 at 15:10
• Ok. Thanks for that, I accepted your answer, it was very helpful. Just curious, can I only add things like \small, \huge etc? Are there no commands that allow actual number specification of the text size? Dec 8 '13 at 15:16
• @user42388 Thank you for accepting! Yes, but it is much easier to go with these commands... Dec 8 '13 at 15:16
• Could you please show me how I can use a number specification? E.g. some of my other headers are in 11 pt, but the reference title looks slightly larger than 11 pt, any ideas? Dec 8 '13 at 15:18


text:
# Use taylor's THeorem to determine the accuracy

1. Apr 29, 2009

### razored

Use taylor's THeorem to determine the accuracy of the approximation: cos(.3) ~=1 - (.3)^2 / 2! + (.3)^4 / 4! when i use taylors theorem, i use (.3)^4 / 4! which gets me 2.03e-10 but the asnwer is R<=2.03e-5

2. Apr 29, 2009

### Dick

Re: Series

(.3)^4/4! is not equal to 2.03e-10. What form of the remainder term are you using?

3. Apr 29, 2009

### razored

Re: Series

Okay, it should be (.3)^5 / 5!

Last edited: Apr 29, 2009
4. Apr 29, 2009

### razored

Re: Series

But i still dont get it, why did i get hte right answer and ignore the f^(n+1)(c) of the formula? (Lagrange error bound) The derivative should have been cos , so i must have picked cos 0. Why?

5. Apr 30, 2009

### HallsofIvy

Staff Emeritus
Re: Series

The error does NOT involve f^(n+1)(c). It involves f^(n+1)(x) for some number between 0 and c (here .3). Since you don't know that number, you use the largest possible value of f^(n+1)(x) between 0 and c to get an upper bound on the error. Since cosine is decreasing, its largest value is cos(0)= 1.

format: plain


text:
# zbMATH — the first resource for mathematics

Differentiable structures on fractal like sets, determined by intrinsic scaling functions on dual Cantor sets. (English) Zbl 0707.58038
Nonlinear evolution and chaotic phenomena, Proc. NATO ASI, Noto/Italy 1987, NATO ASI Ser., Ser. B 176, 101-110 (1988).
[For the entire collection see Zbl 0706.00019.]
This major contribution towards solving the mystery of Feigenbaum universality presents essential ideas in a splendid way, leaving details to the reader.
Consider an embedding of the Cantor set $$C=\{0,1\}^ N$$ into $${\mathbb{R}}$$ obtained by starting with intervals $$I_ 0$$, $$I_ 1$$ and for each word w of zeros and ones dividing $$I_ w$$ into two subintervals $$I_{w0},I_{w1}$$ and a gap $$G_ w$$. The numbers length $$I_{w0}/$$ length $$I_ w$$, length $$mI_{w1}/length I_ w$$ and length $$G_ w/$$ length $$I_ w$$ are said to constitute the ratio geometry of C. They are assumed to be bounded away from zero. Two ratio geometries are considered to be equal if their difference decreases exponentially in length w. The concept of ratio geometry is equivalent to that of a differentiable structure on C - a set of (local) embeddings from C into $${\mathbb{R}}$$ with transition functions in $$C^{1+\epsilon}$$ for some $$\epsilon >0$$. Moreover, the shift map $$J(i_ 0i_ 1i_ 2...)=i_ 1i_ 2..$$. on C is in $$C^{1+\epsilon}$$ for some $$\epsilon$$ if and only if the ratio geometry has a limiting “scaling function” on the dual Cantor set of all left-infinite 0-1 words which is Hölder continuous.
This theory is applied to Feigenbaum-type attractors C of quadratic-like mappings f. Existence of a scaling function for C is equivalent to convergence of the renormalizations of f. Two functions with the same renormalization limit have $$C^{1+\epsilon}$$ conjugate attractors.

##### MSC:
 37D45 Strange attractors, chaotic dynamics of systems with hyperbolic behavior 37E99 Low-dimensional dynamical systems 54H20 Topological dynamics (MSC2010) 37B99 Topological dynamics 37G99 Local and nonlocal bifurcation theory for dynamical systems


text: Tuesday, January 31, 2006 4:30 pm 5130 Upson Hall Ioana Dumitriu University of California, Berkeley Toward accurate polynomial evaluation in rounded arithmetic Given a multivariate (real or complex) polynomial $p$ and a (real or complex) domain $\mathcal{D}$, we would like to decide if there is an algorithm that can evaluate $p$ accurately (i.e., with relative error less than $1$), using rounded (real or complex) "traditional model" arithmetic. We consider this problem both in the classical setting (where the operations are $+$, $-$, $\times$) and in a black-box setting (where other polynomial operations are allowed). We obtain necessary and sufficient conditions for $p$ to be accurately evaluable on $\mathcal{D}$ when $\mathcal{D} = \mathbb{R}^n$ or $\mathbb{C}^n$, and also for some smaller domains. I will also indicate progress toward constructing a complete decision procedure.


text:
Home How do I parse command line arguments in Bash?

# How do I parse command line arguments in Bash?

user6387
1#
user6387 Published in September 21, 2018, 8:03 am

Say, I have a script that gets called with this line:

./myscript -vfd ./foo/bar/someFile -o /fizz/someOtherFile

or this one:

./myscript -v -f -d -o /fizz/someOtherFile ./foo/bar/someFile

What's the accepted way of parsing this such that in each case (or some combination of the two) $v, $f, and $d will all be set to true and $outFile will be equal to /fizz/someOtherFile ?

• For zsh-users there's a great builtin called zparseopts which can do: zparseopts -D -E -M -- d=debug -debug=d And have both -d and --debug in the $debug array echo$+debug[1] will return 0 or 1 if one of those are used. Ref: zsh.org/mla/users/2011/msg00350.html – dezza Aug 2 '16 at 2:13


text:
Session 120 - The Local Interstellar Medium.
Oral session, Thursday, January 18
Corte Real, Hilton

## [120.01] Evidence for Anisotropic Radio Scattering in the Interstellar Medium

A. S. Trotter, J. M. Moran (CfA), L. F. Rodr\'\iguez (UNAM)

The extragalactic radio source NGC 6334B is the most strongly scattered object known, with an angular size of \theta \approx 3\arcsec at 20 cm. Its angular size scales as the square of the observing wavelength, \theta \propto \lambda^2, and its phase structure function, as measured by interferometric visibilities, has a power-law dependence on baseline. Both of these properties are expected if NGC 6334B is being scatter-broadened by an intervening screen of turbulent interstellar plasma. The line of sight to the source intercepts the northern lobe of the bipolar outflow NGC 6334A, which is most likely the source of the scattering. We present observations of NGC 6334B with the VLA at 1.4, 2, 6, 20 and 90 cm, and with the VLBA at 1.4, 2, 3.6 and 6 cm. We performed anisotropic model fits to visibility amplitudes and to closure amplitudes. All the data are consistent with a scattering screen model for this source. The disk is heavily resolved on all but the shortest VLBA baselines, and was undected by the VLA at 90 cm, where the scattering should be particularly severe (\theta \sim 10\arcmin). Moreover, the VLA observations give strong indication that the scattering disk of NGC 6334B is anisotropic, with an axial ratio of \sim 1.2 at 6 and 20 cm, and \sim 1.7 at 2 cm. This anisotropy is most likely due to the effects of a coherent magnetic field permeating the scattering medium. Our fits imply a power-law index in the turbulent density fluctuation power spectrum of 3.7 < \alpha < 4.0, where \alpha = 11/3 corresponds to Kolmogorov turbulence, and \alpha=4 is expected if the interferometer baselines are shorter than the inner (dissipative) scale of turbulence. Measurements of nearby high velocity (80 km\:s^-1) H_2O maser features showed no sign of scatter broadening, implying that they are most likely physically associated with the scattering region.


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# What is the principal argument of $\left( { - 1, - i} \right)$, where $i = \sqrt { - 1}$?$(a){\text{ }}\dfrac{\pi }{4} \\ (b){\text{ }}\dfrac{{ - \pi }}{4} \\ (c){\text{ }}\dfrac{{ - 3\pi }}{4} \\ (d){\text{ }}\dfrac{{3\pi }}{4} \\$

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Hint: In this question we have to find the principal argument of the given coordinates. Now a complex number z can be written in the form of $x + iy$ where x is the real part and y is the imaginary part. Now the argument of any complex number z is ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$. Use this concept to find the principal argument. Keep in mind that here we are asked about the principal argument and not any general argument, so solve accordingly.

We have to find out the principal argument of $\left( { - 1, - i} \right)$.
Now as we know the argument of $\left( {x + iy} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
But in the given coordinate both x and y are negative and we know x and y are both negative in the third quadrant. So, the principal argument of the given coordinate lies in the third quadrant.
Now first calculate the argument of given coordinate so, on comparing,
Y = -1 and x = -1
So, the argument of $\left( { - 1, - i} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{ - 1}}} \right) = {\tan ^{ - 1}}\left( 1 \right)$
Now as we know that the value of ${\tan ^{ - 1}}\left( 1 \right)$ is $\dfrac{\pi }{4}$ in the first quadrant.
And the value of ${\tan ^{ - 1}}\left( 1 \right)$ in third quadrant is $\left[ {\left( {\dfrac{\pi }{4} - \pi } \right) = \dfrac{{ - 3\pi }}{4}} \right]$ or $\left[ {\left( {\pi + \dfrac{\pi }{4}} \right) = \dfrac{{5\pi }}{4}} \right]$
So, the principal argument of $\left( { - 1, - i} \right)$ is $\dfrac{{ - 3\pi }}{4}$ or $\dfrac{{5\pi }}{4}$.
But in options $\dfrac{{5\pi }}{4}$ is not there.
So, the principal argument according to options is $\dfrac{{ - 3\pi }}{4}$.
Hence option (c) is correct.

Note: The note point here is we are asked to find the principal argument and not any general argument. The principal argument is a unique value of argument which always lies in the range $- \pi < {\text{argz < }}\pi$, however a general argument is such that it repeats itself after a regular interval and is given as $2n\pi + {\text{principal argument}}$ where n is an integer.


text:
mersenneforum.org Advice for large GNFS jobs?
 Register FAQ Search Today's Posts Mark Forums Read

2013-07-23, 02:18   #45
swellman

Jun 2012

2·3·479 Posts

Quote:
 Originally Posted by lorgix I'm no expert, but my experiments indicate that you could get away with using r/alim of 2^26-2 and Code: lpbr: 30 lpba: 30 mfbr: 60 mfba: 60 rlambda: 2.7 alambda: 2.7 if you sieve over alim/3 through alim. If that is not a good idea, for whatever reason; let me know.
Again I thank you for your insights. I did a lot more test sieving, focusing on r/alim of 67M. It did improve performance. The best case is the set of parameters you suggested, which gives the minimum estimated sieving time and a yield of about 1.57 rels/spec_q.

All looks good but for one thing - for lpba/r of 30, I need ~93M rels which equates to ~60M spec_q. If I am to run spec_q from alim/3 to alim, I need alim of about 90M.

Or am I missing something? Can I safely run spec_q to values far greater than alim?

For laughs I did run some trials using the 16e siever - great yields but terribly slow. I take it GNFS of 169 is too small for this siever. What's the minimum, 180?

2013-07-23, 07:10   #46
lorgix

Sep 2010
Scandinavia

3×5×41 Posts

Quote:
 Originally Posted by WraithX Hmmm, I've never tried to compile ggnfs before. Can it be compiled in MinGW64? I'm using the binaries that had the Windows ASM improvements provided by Dan Ee over in this thread. Actually, looking at it again, he used MinGW64 and gives instructions on how to compile it. What needs to be changed to allow lpbr/lpba > 33? Also, do you have any thoughts on my question of how to compare the timing (sec/rel) between all my test sieves? Or anyone else, what is the best way to compare these timings? Should I scale the timings the same way I scaled the relations and then perhaps add them up to do a comparison? Or should I just add up the numbers as they are to compare them?
I don't actually know what needs to be changed, or how to do it. But IIRC someone on here said it shouldn't be very hard. Something is limited to 32 bits and it doesn't have to be.
I don't know what the best way to compare timings is. I would certainly scale them to account for the fact that larger lpb needs more rels.
Quote:
 Originally Posted by swellman Again I thank you for your insights. I did a lot more test sieving, focusing on r/alim of 67M. It did improve performance. The best case is the set of parameters you suggested, which gives the minimum estimated sieving time and a yield of about 1.57 rels/spec_q. All looks good but for one thing - for lpba/r of 30, I need ~93M rels which equates to ~60M spec_q. If I am to run spec_q from alim/3 to alim, I need alim of about 90M. Or am I missing something? Can I safely run spec_q to values far greater than alim? For laughs I did run some trials using the 16e siever - great yields but terribly slow. I take it GNFS of 169 is too small for this siever. What's the minimum, 180?
I'm sorry, I may have misled you. I thought 67M would be enough. I am kind of a beginner at this still.
I did not think the yield would be that low. My experiments indicated that the yield would be sufficient. I think you need to raise lpba (and also lim) to get a better yield.
I have just assumed that I can not sieve above lim. That could be wrong.
From what I've seen the cut-off for the 16e-siever is about 180.

If I were you I'd raise lpba and try to figure out which region sieves best.

I hope I didn't waste too much of your time. At least now we know for sure 67M is too low.

 2013-07-23, 07:23 #47 VBCurtis     "Curtis" Feb 2005 Riverside, CA 26·3·23 Posts It is my (beginner) understanding that we can sieve above alim, but yield drops off; if we go too far, we find precious few unduplicated relations. If you look at Rich's parameters in the 4708 thread in aliquot forum (link: http://mersenneforum.org/showpost.ph...postcount=2124), you'll see he has alim at 20M, but plans to sieve from 10M to 34M. I'm not sure why he isn't sieving 7-10M, but his plan indicates it's no big deal to sieve even 40% above alim. -Curtis Last fiddled with by VBCurtis on 2013-07-23 at 07:24 Reason: added link
 2013-07-23, 08:33 #48 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 2×29×109 Posts I find that changing alim doesn't change yield-per-relation very much. Sieving above alim is really not an issue, but once you're at 2.5x or 3x alim the yield does start to drop off. Duplication rate is mostly a function of the siever used, but it's obviously a bit difficult to measure in small samples. I used 16e for a 197-digit gnfs and for a 947-bit snfs, I think that's about the middle of its useful range. For C180 or the 249-digit SNFS I'm doing on and off, I'd be on 15e
 2013-07-23, 08:47 #49 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 5×31×37 Posts In the 64-bit win sievers to remove the 33-bit limit change the 1 to a 0 in the following code snippet: Code:  #if 1 if(max_primebits[i]> 33) { complain("Only large primes up to 33 bits are allowed.\n"); } #endif
2013-07-23, 09:00   #50
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

11000101100102 Posts

Quote:
 Originally Posted by swellman ... for lpba/r of 30, I need ~93M rels ...
That seems rather a lot of relations to aim for, though I suppose it's a C169; for C150-scale numbers, I aim for 80M raw relations and that's usually enough.

I would always aim on the large side for lpba/r; you don't need twice as many relations for one more bit of large primes (I'd expect 150M usually to suffice at 31-bits, and 250M for 32-bits), particularly with the new faster LA step making big matrices less scary. For a C169 I would definitely use 31-bit lpba/r.

2013-07-23, 09:07   #51
lorgix

Sep 2010
Scandinavia

3×5×41 Posts

Quote:
 Originally Posted by VBCurtis It is my (beginner) understanding that we can sieve above alim, but yield drops off; if we go too far, we find precious few unduplicated relations. If you look at Rich's parameters in the 4708 thread in aliquot forum (link: http://mersenneforum.org/showpost.ph...postcount=2124), you'll see he has alim at 20M, but plans to sieve from 10M to 34M. I'm not sure why he isn't sieving 7-10M, but his plan indicates it's no big deal to sieve even 40% above alim. -Curtis
Quote:
 Originally Posted by fivemack I find that changing alim doesn't change yield-per-relation very much. Sieving above alim is really not an issue, but once you're at 2.5x or 3x alim the yield does start to drop off. Duplication rate is mostly a function of the siever used, but it's obviously a bit difficult to measure in small samples. I used 16e for a 197-digit gnfs and for a 947-bit snfs, I think that's about the middle of its useful range. For C180 or the 249-digit SNFS I'm doing on and off, I'd be on 15e
Thanks, I'm learning a lot here.

How should we be choosing alim then? Or better yet; what is alim? And what are the effects from changing it?

2013-07-23, 12:16   #52
swellman

Jun 2012

54728 Posts

Quote:
 Originally Posted by swellman Have you checked Kamada's site? May be helpful on parameter selection (or not). http://homepage2.nifty.com/m_kamada/math/graphs.htm
I've used the models given on this site for several GNFS jobs in the 150s and low 160s with great success. The site collects data from successful factorizations. I will make the observation that successful =/ optimal.

The parameters for composite of GNFS difficulty d are given by

r/alim = INT(3*10^(d/37+3))

lpbr/a = INT(d/18+21)

mfbr/a = INT(d/5+28)

r/alambda = INT(d/26+21)/10

Pure black box approach to be sure, but a sanity check if nothing else.

Similar formulas are given for SNFS.

 2013-07-23, 17:24 #53 firejuggler     Apr 2010 Over the rainbow 46128 Posts just a piece of advice : a line like Code: msieve151_gpu -v -np1 "stage1_norm=3e24 stage2_norm=7e23 800000,1000000" -nps -t 3 work and severly limit you .ms size. -npr after that is much faster, even if you don't put bond. Last fiddled with by firejuggler on 2013-07-23 at 17:24
2013-07-23, 19:04   #54
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

26×3×23 Posts

Quote:
 Originally Posted by firejuggler just a piece of advice : a line like Code: msieve151_gpu -v -np1 "stage1_norm=3e24 stage2_norm=7e23 800000,1000000" -nps -t 3 work and severly limit you .ms size. -npr after that is much faster, even if you don't put bond.
Restricting stage 1 norm is not helpful for poly quality, and composite sizes that produce more stage 1 output by using threads already produce too much for the -nps step to keep up with. I think all -t 3 does for searches that do -nps at the same time is create more GPU heat.

That said, restricting stage 2 norm in the -nps step seems quite useful- my baseline is to divide the default msieve norm by 18-20. I aim for 200-400 hits per day to do -npr on, but tighter -nps stage2norm may be called for.

 2013-07-23, 19:14 #55 firejuggler     Apr 2010 Over the rainbow 244210 Posts what I meant is " don't forget to put a arsh stage2_norm if you use np1 and nps at the same time" for the composite I work on (the last in the poly thread), it seems that any poly of value has a stage2_norm below 1e+023. If it is higher, it won't produce a *good* poly.

 Similar Threads Thread Thread Starter Forum Replies Last Post ryanp Factoring 69 2013-04-30 00:28 ixfd64 Factoring 3 2012-06-06 08:27 WraithX Msieve 18 2012-05-20 22:19 Syd Aliquot Sequences 7 2011-03-14 18:35 bdodson Factoring 20 2008-11-26 20:45

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# Find the difference between radii of smaller and larger circles, whose centers lie on the circle ${x^2} + {y^2} + 4x - 6y - 23 = 0$ and pass through the point (5, 12).A)8B)9C)10D)11E)12

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Hint: General equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, with center of the circle lies on (-g, -f) and radius of the circle is equal to $\sqrt {{g^2} + {f^2} - c}$.

The given circle equation is $${x^2} + {y^2} + 4x - 6y - 23 = 0$$.

Let R be the radius of the above circle.

Comparing the given equation with general equation of the circle we will get g = 2, f = -3, c = -23

$\Rightarrow$Center of the circle (-g, -f) = (2, -3)

$\Rightarrow$ The radius of the circle = $\sqrt {{g^2} + {f^2} - c} = \sqrt {{2^2} + {{( - 3)}^2} - ( - 23)} = \sqrt {36}$

$\Rightarrow$ Radius of the circle R = 6 units.

The geometrical representation of the circle looks similar to the below figure with center at (2, -3) and radius R = 6 units.

Now, we have to draw two circles from the point (5, 12) on a condition that those two circles must have a center lying on the circle $${x^2} + {y^2} + 4x - 6y - 23 = 0$$. And one circle with minimum possible radius and another circle with maximum possible radius satisfying the above condition.
When we draw those two circles it will be similar to the below figure.

[Black circle is the circle with equation$${x^2} + {y^2} + 4x - 6y - 23 = 0$$.

Green color circle represents the smallest possible circle passing through (5, 12) point and its center lies on the black circle.

Red color circle represents the largest possible circle passing through (5, 12) point and its center lies on the black circle.]

We have to find the difference between radii of those largest and smallest circles.

If we observe the figure above it is clear that the difference in radii of green and red circles is equal to the diameter of the black circle.

$\Rightarrow$ Difference = 2R = 2(6) = 12 units.

Option E is the correct answer.

$\therefore$ The difference between radii of smaller and larger circles, whose centers lie on the circle ${x^2} + {y^2} + 4x - 6y - 23 = 0$ and pass through the point (5, 12) = 12 units

Note: Whenever we encounter these kinds of problems we first have to understand the question clearly. Then we have to picturize the given information to get the logic behind it, so that we can easily solve the problem.


text:
# Particle in an Electric Field

1. Jan 18, 2009

### Gogsey

A proton is initially (at time zero) at rest at the origin when a uniform, sinusoidally-varying electric field is turned on, exerting a force F(t) = Fo sinwt on the proton. Calculate the velocity and position as functions of time. Show that the proton moves farther and farther from the origin as time goes on. Calculate the numerical value of the average velocity (averaged over a whole number of cycles) if Fo = eE , w = 2 pi f , with E0 = 5×104 volts/metre, f = 20 MHz.

Firstly there is no velocity term in the equation so I was a little confused, but tried integrating mdv = Fo sinwt dt, and got, V = -Fo coswt/wm, and then again to get x = -Fo sin wt/w^2m.

Can anyone confirm this is right nor not. And if not, what am I doing wrong?

As you can tell, my mechanics is extremely rusty. Thats cause I haven't done any in 2 years, and there was no review or older material at the beginning of this course.

Thanks

2. Jan 19, 2009

### gabbagabbahey

Shouldn't there be a constant of integration or two? Given that v(0)=x(0)=0, what must those constants be?

3. Jan 19, 2009

### Gogsey

Well, those constants should be Vo and Xo, but both are zero at time = 0, so they don't contribute to the question at all.

4. Jan 19, 2009

### gabbagabbahey

Nope, you definitely need to consider the constants of integration. Without them you have:

$$v(t)=\frac{-F_o}{m\omega}\cos\omega t \implies v(0)=\frac{-F_o}{m\omega}\cos(0)=\frac{-F_o}{m\omega}\neq 0$$

and x(t) simply oscillates instead of increasing as expected.

So clearly something is wrong

5. Jan 19, 2009

### Gogsey

Is this anything to do with phase angles? Like cos(wt + phi), and that sort of stuff?

6. Jan 19, 2009

### Gogsey

Can you find the constants of integration by solving the initial value conditions? This seems like the logical thing to do.

7. Jan 19, 2009

### gabbagabbahey

Yes, that's exactly what you need to do. Give it a shot and see what you get.

8. Jan 19, 2009

### Gogsey

Ok, so the first one should be Fo/mw. The next ones a little trickier. First attempt I got, the constant to be zero for the seond integral.

9. Jan 19, 2009

### gabbagabbahey

Good

Better show me your attempt at that one

10. Jan 19, 2009

### Gogsey

Ok,

So for v(t)=-Fo sinw(t)/mw + Fo/mw.

Actually, th prof skipped the integration for finding X, although it was an equation without any constant, ie, this equation had a cos function, so there was no constant in this part. It was simply dv/dt=Fo sinw(t), and he skipped the integration, to find x. So I don't really know how to do this.

All I can think of is that it could be x=Fo sinw(t)/mw^2 + Fot/mw + C. And since t is zero,
C=0.

Last edited: Jan 19, 2009
11. Jan 19, 2009

### gabbagabbahey

You mean x(t) right?

$$v(t)=\frac{-F_0}{m\omega}\cos\omega t +\frac{F_0}{m\omega} \implies x(t)=\int v(t)dt=\frac{-F_0}{m\omega}\sin\omega t +\frac{F_0 t}{m\omega}+C$$

When you substitute the initial conditions, you should find that $C=0$.

What happens to x(t) as $t \to \infty$?

To average some periodic function f(t) over a period T, you use the formula $$f_{ave}=\frac{1}{T}\int_0^T f(t)dt$$

Apply that to v(t) to find your average velocity.

12. Jan 19, 2009

### Gogsey

Sorry, a bit lost now, I haven't seen this before, at least that I can remeber. Definatley not seen it in my 2nd year Mechanics course.

13. Jan 19, 2009

### gabbagabbahey

You should have covered time averages in Calc I or II. Try applying the formula and see what you get.

14. Jan 19, 2009

### Gogsey

Does this mean that I inetegrate v(t), and then after the integration I get what I had before times 1/t, then c = this integral evaluated at t(cause at zero its zero), times T?

15. Jan 19, 2009

### gabbagabbahey

This integral has nothing to do with determining $C$. You already showed that $C=0$.

This integral is what you need to compute in order to find the average velocity and answer this part of your question:

In the formula I gave above, capital T represents the period of v(t).

$$T=\frac{1}{f}=\frac{2\pi}{\omega}$$

16. Jan 20, 2009

### Gogsey

This may sound stupid, and it shows how little I know about this, but when you find Vav, which turns out to be just Xo (fromT to 0) * 1/T, which is then is just x(T) * 1/T.

Do you sub in T for the sin wt part, ie for the w(t)?

I'm pretty sure you don't but I wouldn't want to make this silly mistake.

17. Jan 20, 2009

### gabbagabbahey

Of course. To find the value of x(T), you substitute t=T everywhere into the expression for x(t).

18. Jan 20, 2009

### Gogsey

Maybe I said that wrong.

Vav = 1/T (Fosinwt/mw^2 + FoT/mw)

So 1/T cancles with the second term T to get Fo/mw.

Now should the first term be Fo sin wt/Tmw^2, or Fo sin wT/TmW^2?

19. Jan 20, 2009

### gabbagabbahey

Fo sin wT/TmW^2

for the exact reason stated in my previous post.

20. Jan 20, 2009

### Gogsey

Awesome., thanks. And wT still is 2 pi f, which is given in the equation, right? By going from wt to wT, nothing has changed in this sense.


text:
# Cayley's formula

The complete list of all trees on 2,3,4 labeled vertices: $2^{2-2}=1$ tree with 2 vertices, $3^{3-2}=3$ trees with 3 vertices and $4^{4-2}=16$ trees with 4 vertices.

In mathematics, Cayley's formula is a result in graph theory named after Arthur Cayley. It states that for any positive integer n, the number of trees on n labeled vertices is $n^{n-2}$.

The formula equivalently counts the number of spanning trees of a complete graph with labeled vertices (sequence A000272 in OEIS).

## Proof

Many remarkable proofs of Cayley's tree formula are known.[1] One classical proof of the formula uses Kirchhoff's matrix tree theorem. Prüfer sequences yield a bijective proof of Cayley's formula. Another bijective proof, by André Joyal, finds a one-to-one transformation between n-node trees with two distinguished nodes and maximal directed pseudoforests. A proof by double counting due to Jim Pitman applies to trees.

## History

The formula was first discovered by Carl Wilhelm Borchardt in 1860, and proved via a determinant.[2] In a short 1889 note, Cayley extended the formula in several directions, by taking into account the degrees of the vertices.[3] Although he referred to Borchardt's original paper, the name "Cayley's formula" became standard in the field.


text:
# Prove there are no rational solutions

• Apr 18th 2010, 02:49 PM
treetheta
Prove there are no rational solutions
Prove that there are no rational solutions to the equation x^3 + x^2 = 1
• Apr 18th 2010, 04:19 PM
Laurent
Quote:

Originally Posted by treetheta
Prove that there are no rational solutions to the equation x^3 + x^2 = 1

Let $x=\frac{p}{q}$ with $p\wedge q=1$; you get $p^3+qp^2=q^3$ hence (...) $q|p$ and thus $q=1$, which mean $x$ is an integer. I let you fill in the (quick) dots and handle it from there (beware, $x^3$ may be negative).
• Apr 18th 2010, 07:17 PM
Drexel28
Quote:

Originally Posted by treetheta
Prove that there are no rational solutions to the equation x^3 + x^2 = 1

Or, maybe the rational root theorem...


text:
# 2015 AMC 10B Problems/Problem 22

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Solution

Triangle $AFG$ is isosceles so $AG$=$AF$=$1$. Using the symmetry of pentagon $FGHIJ$, notice that $\angle{FAG}$ is congruent to $\angle{FBI}$, so triangles $JHG$ and $DIJ$ are congruent to our original triangle $AFG$. Therefore, $JH$ = $1$. Now, we still need to find the length of $FG$ and $DC$. Also, we know that $FJ$ = $FG$ since pentagon $FGHIJ$ is regular. Let's call the length of $FG$ and $FJ$ $x$. Now we can solve for $x$. Triangles $AFG$ and $AJH$ are similar. So,

$$\frac{1}{x+1} = \frac{x}{1}$$

From this, we get $x = \frac{\sqrt{5} -1}{2}$.

Now, we just have to find the length of $DC$ which we'll call $y$. We already know that $DJ$ = $AF$ = $1$. Triangles $AFG$ and $ADC$ are similar so we have,

$$\frac{1}{2+x} = \frac{x}{y}$$

Solving for $y$ we get $y = \frac{\sqrt{5} +1}{2}$

Adding up $FG$ , $JH$, and $CD$ gives us $1 + x + y$ which is $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Solution by arebei2


text:
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# Which of the following inequalities is the solution to the inequality

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Which of the following inequalities is the solution to the inequality [#permalink]

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Which of the following inequalities is the solution to the inequality 7x – 5 < 12x + 18?

(A) x < –13/5
(B) x > –23/5
(C) x < –23/5
(D) x > 23/5
(E) x < 23/5
[Reveal] Spoiler: OA

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Kudos [?]: 347 [0], given: 594

Which of the following inequalities is the solution to the inequality [#permalink]

### Show Tags

05 Nov 2017, 13:10
Bunuel wrote:
Which of the following inequalities is the solution to the inequality 7x – 5 < 12x + 18?

(A) x < –13/5
(B) x > –23/5
(C) x < –23/5
(D) x > 23/5
(E) x < 23/5

$$7x – 5 < 12x + 18$$

$$-23 < 5x$$

$$-\frac{23}{5} < x$$

$$x > -\frac{23}{5}$$

Kudos [?]: 347 [0], given: 594

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Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1713

Kudos [?]: 914 [0], given: 5

Re: Which of the following inequalities is the solution to the inequality [#permalink]

### Show Tags

08 Nov 2017, 17:34
Bunuel wrote:
Which of the following inequalities is the solution to the inequality 7x – 5 < 12x + 18?

(A) x < –13/5
(B) x > –23/5
(C) x < –23/5
(D) x > 23/5
(E) x < 23/5

We can simplify the given inequality:

7x - 5 < 12x + 18

-23 < 5x

-23/5 < x

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 914 [0], given: 5

Re: Which of the following inequalities is the solution to the inequality   [#permalink] 08 Nov 2017, 17:34
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text:
• ### Background model for the NaI(Tl) crystals in COSINE-100(1804.05167)

April 14, 2018 physics.ins-det, astro-ph.IM
The COSINE-100 dark matter search experiment is an array of NaI(Tl) crystal detectors located in the Yangyang Underground Laboratory (Y2L). To understand measured backgrounds in the NaI(Tl) crystals we have performed Monte Carlo simulations using the Geant4 toolkit and developed background models for each crystal that consider contributions from both internal and external sources, including cosmogenic nuclides. The background models are based on comparisons of measurement data with Monte Carlo simulations that are guided by a campaign of material assays and are used to evaluate backgrounds and identify their sources. The average background level for the six crystals (70 kg total mass) that are studied is 3.5 counts/day/keV/kg in the (2-6) keV energy interval. The dominant contributors in this energy region are found to be $^{210}$Pb and $^3$H.
• ### Initial Performance of the COSINE-100 Experiment(1710.05299)

COSINE is a dark matter search experiment based on an array of low background NaI(Tl) crystals located at the Yangyang underground laboratory. The assembly of COSINE-100 was completed in the summer of 2016 and the detector is currently collecting physics quality data aimed at reproducing the DAMA/LIBRA experiment that reported an annual modulation signal. Stable operation has been achieved and will continue for at least two years. Here, we describe the design of COSINE-100, including the shielding arrangement, the configuration of the NaI(Tl) crystal detection elements, the veto systems, and the associated operational systems, and we show the current performance of the experiment.
• ### Muon detector for the COSINE-100 experiment(1712.02011)

The COSINE-100 dark matter search experiment has started taking physics data with the goal of performing an independent measurement of the annual modulation signal observed by DAMA/LIBRA. A muon detector was constructed by using plastic scintillator panels in the outermost layer of the shield surrounding the COSINE-100 detector. It is used to detect cosmic ray muons in order to understand the impact of the muon annual modulation on dark matter analysis. Assembly and initial performance test of each module have been performed at a ground laboratory. The installation of the detector in Yangyang Underground Laboratory (Y2L) was completed in the summer of 2016. Using three months of data, the muon underground flux was measured to be 328 $\pm$ 1(stat.)$\pm$ 10(syst.) muons/m$^2$/day. In this report, the assembly of the muon detector and the results from the analysis are presented.


text:
# Uniformization of embedded surfaces

by Neil Strickland, and the related Maple code.

Let $X\subset S^3$ be a smoothly embedded closed surface of genus $g\gt 1$. As we will explain, $X$ automatically has a very rich and rigid geometric structure. Indeed, $X$ inherits a Riemannian metric from $S^3$, and after specifying some conventions we also obtain a well-defined orientation. Now consider a point $x\in X$, and let $T_xX$ denote the corresponding tangent space. Let $J_x\colon T_xX\to T_xX$ be the anticlockwise rotation through $\pi/2$ (which is meaningful given the metric and orientation). This depends smoothly on $x$ and satisfies $J_x^2=-1$, so it gives an almost complex structure on $X$. It has been known since the early twentieth century that any almost complex structure on a manifold of real dimension two integrates to give a genuine complex structure. Thus, $X$ can be regarded as a compact Riemann surface. It is known that any compact Riemann surface can be regarded as a projective algebraic variety over $\mathbb{C}$, and also as a branched cover of the Riemann sphere. Alternatively, as we have assumed that the genus is larger than one, the universal cover of $X$ is conformally equivalent to the open unit disc $\Delta$. This means that $X$ is conformally equivalent to the quotient $\Delta/\Pi$ for some Fuchsian group $\Pi$.

To the best of our knowledge, the literature contains no examples where a significant fraction of this structure can be made explicit. This project is a partially successful attempt to provide such an example, involving the surface $EX^* = \{x\in S^3\; | \; (3x_3^2-2)x_4+\sqrt{2}(x_1^2-x_2^2)x_3=0 \}$

• The monograph itself is on the arxiv and also here.
• There is also a large body of related Maple code and worksheets, together with graphics and data files produced by that code.
• The arxiv page also has the code and worksheet files, uploaded as ancilliary files with the main LaTeX document. However, readers may find it more convenient to use this page instead of the arxiv page.
• As well as the discussion of the Maple code structure in the memoir, there are some documentation files:
• The code and worksheets can be downloaded as a .zip file or as a .tgz file. These files contain only a few of the graphics generated by the code, and none of the data files. To start using the code, one should open the file genus2.mw in the worksheets directory.
• Alternatively, the above files can be obtained from the GitHub repository.
• One can also download much larger files containing all graphics and data as well as the code, in .zip format (300MB) or .tgz format (190MB).


text:
# [texhax] Greek letters with superscripts in list environment

Sam Albers tonightsthenight at gmail.com
Tue Dec 2 03:58:02 CET 2008

Hello,

For some reason I can't get LaTex to produce a greek letter with a
supercript while in a list environment. Is there any way around this issue?
Here is part of my file:

\begin{itemize}

\item The ratio of flow depth to grain size (A)

\item the density ratio (B)

\item the Reynolds number (Re)

\item the Froude number (Fr)

*\item shear stress ($\tau$^{*})*

\end{itemize}

This doesn't work. However with the superscript it works fine. Any ideas?

Thanks

Sam
-------------- next part --------------
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URL: http://tug.org/pipermail/texhax/attachments/20081201/4bd5682b/attachment.html


text:
# 原创初中英语自诊断-初中英语重点常考句型汉译英

26. Stop making that noise.别这么吵闹。（stop doing\to do）

We stopped (walking) to have lunch in a village pub.

27.It’s dangerous (for you) to drive so fast. (It’s +adj.+for\of sb to do)

（你）车开得这么快是危险的。

It’s impossible for him to work out this maths problem.

I found it impossible to work out this maths problem. (find it+adj.+to do)

28. He showed me his new suit. / He showed his new suit to me.

She bought a dress for me. She bought me a new dress.

29. She spent a lot of money on clothes. (spend time/money on sth.)

She spent a lot of time (in) surfing. (spend time/money in doing )

He paid $10 for the book.(pay for sth.) The book cost him$10.(sth. cost sb.some money)

30. What do you mean by “baozi”?( what do you mean…)

“baozi”是什么意思？

31. What/How about having a rest? /(having) a try? (What/How about + doing)

32. Why not make friends with him? (Why not do)（why don’t you do）

33. It’s better to make up your mind. (It’s better\best to do)

It’s best to wear silk clothes in summer.

34. The river is 50 kilometres long and two metres deep.

35. Your cake is as delicious as mine.(as…as…)

He doesn’t run so / as fast as you.

36. He often helps me with my Chinese. (help sb. with sth.)

He often helps me do the homework. (help sb. do sth.)

37. What do you think of the novel?( what do you think of/ how do you like)

How do you like the novel?

38. He likes cooking very much. So do I. / So does he.(so+倒装语序)

He is American. So is Jim.

39.Don’t forget to post the letter on your way home.

Did you remember to send your mother a birthday card last week?

I’ll never forget seeing the girl for the first time.

40. Forests help to keep water from running away. (keep\stop\prevent from doing)

The Great Green Wall will stop the sand from moving towards the rich farmland.

41. I prefer to do shopping on the Internet rather than go to the shop.

42. He has nothing to do with the club. (have nothing to do with)

43. Tom made such a noise that his sister told him to be quiet.(such…that…)

His idea was such a good one that we all agreed to use it.

44. We are proud of what you’ve done. (be proud of)

We are proud of our city.

45. Beijing is famous for its many places of interest in the world.( be famous for)

46.The students are getting ready for the coming sports meeting.

47.There is something wrong with my bicycle .can you help me with it?

48.I’ll call you as soon as I arrive in shanghai。( as soon as)

49.She is good at dancing。(be good at ,do well in)

50.They are interested in animals。( be interested in)


text:
# $z_{0}$ is a pole of $f$

Let $f$ be an analytic function on a punctured neighborhood of $x_{0}\in\mathbf{C}$, that is, $f$ analytic on

 $\{z\in C:0<|z-x_{0}|<\varepsilon\}$

for some $\varepsilon>0$ and such that

 $\lim_{z\rightarrow z_{0}}|f(z)|=\infty.$

We say then that $x_{0}$ is a pole for $f$.

Title $z_{0}$ is a pole of $f$ Z0IsAPoleOfF 2013-03-22 14:00:55 2013-03-22 14:00:55 drini (3) drini (3) 7 drini (3) Definition msc 30A99


text:
# On $\#P\subseteq FP^{\Sigma_{f(n)}^P}$?

1. Is it known that permanent of a $0/1$ $n\times n$ matrix $M$ is computable in polynomial or randomized polynomial time with access to a ${\Sigma_{(\log n)^c}^P}$ oracle where $0<c$ holds and $\Sigma_{f(n)}^P$ means the number of alternations is $f(n)$?

2. If not known what is the consequence if it is possible to do so?

This is very unlikely to hold, but (as usual) impossible to rule out using current techniques, as we can’t even prove $\mathit{PSPACE}\ne P$.
However, we can at least show that no relativizing argument proves such an inclusion, i.e., there are oracles $X$ such that randomized polynomial-time with a $(\Sigma^P_{(\log n)^c})^X$ oracle does not include $\#P^X$. Better yet, there are oracles where it does not even include $\oplus P^X$ (which is included in $P^{\#P^X}$): otherwise, there would exist $O((\log\log n)^c)$-depth quasipolynomial-size unbounded fan-in circuits that approximate parity with error, say, $1/4$, contradicting Håstad’s theorem.


text:
# integration--Finding Principal Value

#### clickbb08

##### New member
I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.

#### dwsmith

##### Well-known member
I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.
First it should be
$$\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = \frac{\pi}{3e^3}$$

Let $f(z) = \dfrac{e^{iz}}{(z-3i)(z+3i)}$

Then
$$\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = 2\pi i\sum_{\text{UHP}}\text{Res}_{z=z_j}f(z) + \pi i\sum_{\text{real axis}}\text{Res}_{z=z_j}f(z)$$

UHP = upper half plane

Last edited:


text:
Appearance-based modeling for segmentation of hippocampus and amygdala using multi-contrast MR imaging

No Thumbnail Available
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2011
Authors
Hu, Shiyan
Coupé, Pierrick
Collins, D. Louis
Journal article
Published
Published in
NeuroImage ; 58 (2011), 2. - pp. 549-559. - ISSN 1053-8119. - eISSN 1095-9572
Abstract
A new automatic model-based segmentation scheme that combines level set shape modeling and active appearance modeling (AAM) is presented. Since different MR image contrasts can yield complementary information, multi-contrast images can be incorporated into the active appearance modeling to improve segmentation performance. During active appearance modeling, the weighting of each contrast is optimized to account for the potentially varying contribution of each image while optimizing the model parameters that correspond to the shape and appearance eigen-images in order to minimize the difference between the multi-contrast test images and the ones synthesized from the shape and appearance modeling. As appearance-based modeling techniques are dependent on the initial alignment of training data, we compare (i) linear alignment of whole brain, (ii) linear alignment of a local volume of interest and (iii) non-linear alignment of a local volume of interest. The proposed segmentation scheme can be used to segment human hippocampi (HC) and amygdalae (AG), which have weak intensity contrast with their background in MRI. The experiments demonstrate that non-linear alignment of training data yields the best results and that multimodal segmentation using T1-weighted, T2-weighted and proton density-weighted images yields better segmentation results than any single contrast. In a four-fold cross validation with eighty young normal subjects, the method yields a mean Dice к of 0.87 with intraclass correlation coefficient (ICC) of 0.946 for HC and a mean Dice к of 0.81 with ICC of 0.924 for AG between manual and automatic labels.
150 Psychology
Cite This
ISO 690HU, Shiyan, Pierrick COUPÉ, Jens C. PRUESSNER, D. Louis COLLINS, 2011. Appearance-based modeling for segmentation of hippocampus and amygdala using multi-contrast MR imaging. In: NeuroImage. 58(2), pp. 549-559. ISSN 1053-8119. eISSN 1095-9572. Available under: doi: 10.1016/j.neuroimage.2011.06.054
BibTex
@article{Hu2011-09Appea-38403,
year={2011},
doi={10.1016/j.neuroimage.2011.06.054},
title={Appearance-based modeling for segmentation of hippocampus and amygdala using multi-contrast MR imaging},
number={2},
volume={58},
issn={1053-8119},
journal={NeuroImage},
pages={549--559},
author={Hu, Shiyan and Coupé, Pierrick and Pruessner, Jens C. and Collins, D. Louis}
}

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No


text:
# Invariant Rings of p-groups

• Katherine Horan, University of Kent
• Friday 06 November 2015, 15:00-16:00
• CMS, MR15.

Let $k$ be a field, $V$ a $k$ vector space and $G \leq GL(V)$ a finite group. In invariant theory we are interested in the invariant ring $k[V]G:=\{f \in k[V] |\, g(f) = f, \, \forall g \in G\}$, one of the questions we can ask is when is this polynomial? Are there cases when $k[V]^G$ is not even Cohen Macaulay? I will give a brief introduction to what is known in the modular and non-modular cases, with some more specific results and examples for $p$-groups, $k$ characteristic $p$.

This talk is part of the Junior Algebra/Logic/Number Theory seminar series.


text:
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# assembly help??

This topic is 6173 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I have read assembly turtorials but they are to confusing. Can someone please tell me how assembly works and some of the functions. THANKS

##### Share on other sites
look here http://win32asm.rxsp.com/

example of a program that display an MessageBox (under Win32)
; ################################################      .386                     ; compile for 386 (32 bits)      .model flat, stdcall     ; memory model and call directive (param last to first)      option casemap :none      ; case sensitive; ################################################      include \masm32\include\windows.inc      include \masm32\include\user32.inc      include \masm32\include\kernel32.inc      includelib \masm32\lib\user32.lib      includelib \masm32\lib\kernel32.lib; ################################################    .code      szDlgTitle    db "Minimum MASM",0      szMsg         db "  --- Assembler Pure and Simple ---  ",0start:    invoke MessageBox,0,ADDR szMsg,ADDR szDlgTitle,MB_OK    invoke ExitProcess,0end start

This code take 1.5ko when it is compiled.

Basically asm can just use processor function, like call (call a function), jmp (jump to an instruction), mov (mov memory to another place), add (add to register)....
(in this example invoke is a macro for severall push and a call)

As you can see, asm can call Win32 API, but it can make some complexe things like COM (yes asm can make object)

Why English rules?? C pas très malin tout ça!

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Assembly doesn''t really have "functions", ASM is a low level form of coding by manipulating CPU registers and bios interrupt request to get a certain task done. As for full windows ASM i don''t suggest it, waste of time a couple reasons:

1. Windows doesn''t send Win32 executible instructions directly to the processor, but to the kernel and other residing system dll''s to carry out most tasks I.E., create windows and so forth.

2. There are many languages such as C++ and Visual Basic that can create a Win32 project with just as much speed as an ASM Win32 project, mainly because windows is a system of interruptions as I put it, say you right click on a window, windows then knocks on the door of the program code and says hey this person click the right mouse button whats going to happen. So basically windows programs are behind a wall that catches events to be executed when something happens.

3. Assembly is outdated.

4. For a simple useful Win32 project it can take ages, and many lines of code that is very hard to understand.

Thats all I really have to say about that, I think you should learn ASM simply for the fact of understanding the CPU and how the computer works but for programming a huge task it would be just a big waste of time.

MOV AH,3h
INT 21Ch

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i''m guessing your not going ot check the post

MOV AH,3h
INT 21Ch

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Shouldn''t that be "int 21h" not "int 21ch"?

Also, assembly is never really outdated, as compilers translate your C++ code into assembly, but usually not perfectly. Hand-written assembly allows for slightly better optimization, and most importantly, it allows you to take advantage of feature sets the compiler doesn''t use, such as MMX (PPro+) and XMM (PIII+).

If you were to go about learning assembly, xterminator, I strongly recommend you stick to DOS assembly for the time being... that way you don''t have to worry about using external functions at all; you can learn how to work with files and print text without relying on functions like fopen or printf.

~CGameProgrammer( );

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quote:
Original post by x86ASMRoutine

3. Assembly is outdated.

Not so.

Although assembly is not often used in modern games, it can still be applied in some scenarios. It all depends on what you''re project goals are. Japanese developers still use assembly language in modern console games (Tecmo, creators of the Dead or Alive series, often boast of their highly skilled assembly programmers.) Consoles are still a great playground for assembly language... Even though assembly is becoming less used in modern systems, it still yields benefits on complex systems such as the PS2, which cannot be pushed to the max without assembly (or so I''ve heard from certain developers.)

Nonwithstanding major changes to the concepts of computer architecture, assembly language is never outdated; in fact, it is always farther ahead of anything else simply because it _is_ the machine language.

quote:

Thats all I really have to say about that, I think you should learn ASM simply for the fact of understanding the CPU and how the computer works but for programming a huge task it would be just a big waste of time.

I agree that mastering assembly language _really_ helps out with understanding how computers work. It''s tough for a newcomer in this day and age to pick up on assembly language, simply because so much emphasis is placed on abstraction and high level programming nowadays. The warm world of assembly (where everyone knows your name) becomes a harsh and unforgiving place for beginners.

Assembly is often taught incorrectly. I haven''t seen any good tutorials suitable as an introduction. There are some good books out there, though, but even they aren''t completely sufficient. I started with "The Revolutionary Guide to Assembly Language" from Wrox. It wasn''t a very good book, but it helped a little.

I think Jeff Duntemann''s book would be a better starting place. Unfortunately, I didn''t have it when I was learning X86 assembly.

Another good, but very expensive and rare book is "The 80x86 IBM PC and Compatible Computers (Volumes I & II) Assembly Language, Design, and Interfacing" by Muhammad Ali Mazidi and Janice Gillispie Mazidi. It''s a wonderful book, but as with all assembly books, many of the topics it covers seem archaic -- but definitely worth learning, if you really want to learn about the IBM PC. So far, it''s the clearest book on the topic that I''ve seen. It''s packed with a lot of information on PC hardware interfacing, too.

The important thing to remember is that assembly is really just a bunch of mnemonics for processor instructions. You have a ton of power and flexibility at your disposal. Writing entire PC applications in assembly nowadays would usually be considered insane. Assembly is best used in very frequently executed performance-intensive code. Often times, the assembly implementation can be cleaner than a C implementation... it depends on the situation.

One of the greatest things about mastering an assembly language (especially X86) is that other assembly languages are a snap to learn! I''ve never had problems coding in assembly for PowerPC, 6502, 68K, i960, Z80, etc. A processor reference and an assembler are enough to get me going right away, with pretty much no time spent having to learn the language -- you just learn the mechanics of it as you go along.

IMHO, no high level language can give you such an understanding of the hardware you''re working with. It''s a tough subject to get into, but it''s all worth it once you look around and realize how much knowledge you''ve gained

quote:

MOV AH,3h
INT 21Ch

I''m assuming you meant INT 21h? There are only 256 interrupt vectors on the X86... Typeo

---
Bart

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Thanks I didn''t notice the mistake in my sig =/ All fixed

MOV AH,3h
INT 21h

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## The Origin of Space and Time by Design, not by Formula.

http://vixra.org/author/leo_vuyk
https://www.flickr.com/photos_user.gne?
path=&nsid=93308747%40N05&page=&details=1

If the big bang was the splitting of a huge Axion/ Higgs particle knotty Dark Matter Black Hole (DM- BH) nucleus into smaller DM-BH nuclei, then no standard Fermion/ Baryon inflation has happened only the DM-BH based Lyman alpha forest equipped with local Herbig Haro star/galaxy creating systems.

All black holes of all sizes (down to ball lightning) seem to be equipped with a Fermion repelling- and plasma producing horizon, which has also a charge splitting effect into a negative (outside) and positive ( inside) zone ( see orientale basin of the moon) .

## Sunday, October 30, 2016

### Moon’s gravity anomaly, a support for a new physics Fermion repelling and charge splitting black hole.

Friendly New physics black holes on the moon and all other places in the multiverse..
According to Quantum FFF Theory, comet nuclei are plasma producing new physics black holes with a Fermion repelling horizon around the nucleus.

Crashed comets even seem often to split before  crashing into stars, (forming starspots) planets or moons leaving extra peculiar pensil shaped gravity patterns behind. (see Orientale basin).
https://www.flickr.com/photos/93308747@N05/albums/72157651837150515

## Saturday, October 29, 2016

### Spooky gravity anomalies on earth and moon by Cometary origin, with a new physics Fermion repelling black hole !

According to Quantum FFF Theory, comet nuclei are plasma producing new physics black holes with a Fermion repelling around the nucleus. (see Orientale basin).  Blue colored ring around the red center, seems to represent the Fermion repelling zone around the central large cometary black hole with extra gravity ( red) .
Crashed comets even seem often to split before  crashing into planets or moons leaving extra peculiar gravity patterns behind.

see:

# Insights into giant impacts on moon, Earth and Mars

https://www.sciencedaily.com/releases/2016/10/161027141000.htm

### The origin of Pumpkin stars in critical Herbig Haro systems

According to Quantum FFF Theory, stars are formed between dual new physics black holes called Herbig Haro pressure cookers.
The dual black holes can have different sizes and different capacities to produce different sized stars.
They are assumed to produce most of the the plasma dust and and Hydrogen gas to form the central baby star.
The  the size of both black holes is the origin of the polarization of the oscillating Higgs field vacuum (spacetime structure) in the HH system.
The Higgs field vacuum is assumed to be responsible for the maximum size of the baby star, before instability into a pumpkin star.
Pumpkin stars of different sizes, are assumed to be the start of secondary sibling stars and the start of an open star cluster producing larger or smaller stars and even failed stars or dwarfs.

Pumpkin stars of different sizes, are assumed to be the start of secondary sibling stars and the start of an open star cluster producing larger or smaller stars and even failed stars or dwarfs.see below Bok globule ESO 9934.:

## Thursday, October 27, 2016

### Proton density bubble in double magic Si34

The report of the first experimental evidence that points to a depletion of the central density of protons in the short-lived nucleus 34Si.
http://www.nature.com/nphys/journal/vaop/ncurrent/full/nphys3916.html
The New Nuclear Magic Number (34) Explained by the Polar Coaxial Ring System of Quantum FFF Theory.
http://vixra.org/pdf/1310.0101v3.pdf
and http://vixra.org/pdf/1102.0054v1.pdf
Atomic Nuclear geometry based on Magic Number logic.

### Binary star production out of a mother star, (L1448 IRS3B system) the start of an open star cluster

According to the Quantum FFF model, Young stars can produce new baby by overflow of mass input between Herbig haro new physics black holes.
The size of the black holes seems to be resposnible for the maximum size of the star before it start to cannonball produce baby stars or later failed stars or planets
see: https://www.sciencedaily.com/releases/2016/10/161026133058.htm

## Wednesday, October 26, 2016

### Tabby’s star equipped with “alien megastructures” explained by oversized star spots or new physics micro black holes.

According to the Quantum FFF model, comet nuclei are formed by CHO producing micro electric dark matter black holes.
Crashed comets are assumed to be the origin of sunspots and all other pointlike energetic phenomena in stars planets and moons like volcanos at Jupiters moon IO.
However the gravity and light dimming effect is supposed to be multiplied if two "starspots" are aligned with the line of sight, which I call: pencil shaped lighthouse effect. also found on earth an Saturn rings.

As a consequence, alien magastructure star or Tabby's star (KIC 8462852) could be
interpreted by the existence of over sized starspots or crashed comet remnants.
For the original paper: see: https://arxiv.org/pdf/1608.01316v1.pdf
called: KIC 8462852 FADED THROUGHOUT THE KEPLER MISSION
by Benjamin T. Montet1,2 and Joshua D. Simon
and for the possible Q-FFF model solution see: https://www.flickr.com/photos/93308747@N05/albums/72157633348892116

Are Sunspots made of gravitating Dark Matter Black Holes?
http://vixra.org/pdf/1312.0076v1.pdf
Birkeland Currents, Sunspots, Comets and Ball lightning originated by new paradigm
Dark Matter Black Holes.
http://vixra.org/pdf/1209.0061v2.pdf
Evidence for Dark Matter Black Hole Based Plasma, Dust and Ice Production Inside Comets Like 67P,C-G. and the Growing Earth.
http://vixra.org/pdf/1410.0039v6.pdf

Other examples closer to home.
Temporary spokes at the rings of Saturn, by dual black hole pencil gravity effect.

## Tuesday, October 25, 2016

### The universe is not expanding as we are told.

Hubble Redshift combined with Universal Contraction is possible.
http://vixra.org/pdf/1303.0211v1.pdf
http://bigbang-entanglement.blogspot.nl/2016/07/the-hubble-redshift-is-cheeting-us-at.html
http://motls.blogspot.nl/2016/10/dark-energy-and-unsettled-science.html

Three peaks in the CMB radiation assumed to be the “contact areas” with neighbour universes (one anti-material and two material?)
The red peak area could have more energy due to much more annihilation processes between matter and antimatter, at the contact areas.
Our position in the 8 or 12 !!! fold raspberry multiverse.
Elliptic shaped CMB peak radiations are tell tales. (left).
At the rightside a 12 fold  symmetric cube octahedron multiverse diagram is depicted with red: matter and green: anti matter entangled universes. SEE:
http://arxiv.org/pdf/1512.00554v2.pdf
For WMAP CMB info image, Credit: V. G. Gurzadyan and R. Penrose

# How far away is your parallel self?

see below:
It is well known that not only elements of the standard model combined with Quantum mechanics and relativity rules are needed to get a shimmer of a TOE.
However a TOE should be combined with an experimental verified model for free will.

The Quantum FFF model (Function Follows Form) seems to point to a solution of a raspberry shaped instant entangled Charge Parity  symmetric multiverse.

The raspberry shaped multiverse as a Cubeoctaheron (with 4x elliptic contact areas)? or a Dodecahedron ( with 5x elliptic contact areaas) ? or a cube ( with 3x elliptic contact areas)?
See two articles:
Democratic Free Will in the instant Entangled Multiverse.

It seems to be high time, to do the extended Libet experiment described in this paper.
And:
New physics elements in the Quantum Function Follows Form Model.

Alternative for Dodecahedron: the Cube-octahedron, based on observation of extreme elliptic fluctuations the CMB.
Three peaks in the CMB radiation assumed to be the “contact areas” with neighbour universes (one anti-material and two material?)
The red peak area could have more energy due to much more annihilation processes between matter and antimatter, at the contact areas.
Our position in the 8 or 12 !!! fold raspberry multiverse.  Elliptic shaped CMB peak radiations are tell tales. (left). At the rightside a 12 fold  symmetric cube octahedron multiverse diagram is depicted with red: matter and green: anti matter entangled universes. SEE:
arxiv.org/pdf/1512.00554v2.pdf

For WMAP CMB info image, Credit: V. G. Gurzadyan and R. Penrose

Quantum FFF Theory is also published in the form of POSTERS at the Flickr site: http://www.flickr.com/photos/93308747@N05/

Numbered listing of Vixra essays by Leo Vuyk.

Quantum FFF Theory principles.
Quantum FFF Theory is based on Non-Local Realism and rigid stringy Axion-Higgs particles able to mutate into all other particles.
The CP(T)-symmetrical raspberry shaped multiverse (with 8-12 berries) is non local instant connected down to each quantum. Each quantum can be visualised as a deformed Higgs ring able to convert into other shapes and able to form compound structures like quarks, muons or even Tau leptons. So realism (cogwheels) connected by symmetric non-local instant entanglement.

The basic elements:
1: Black holes (BHs) are the same as Dark Matter, they all consume photons, even gravitons and the Axion-Higgs field, ( for short: Higgs field) but REPEL Fermions due to their propeller shape polarisation away from the BH. They produce electric charged plasma.
2: Dark Energy is the oscillating ( Casimir) energy of the massless Axion-Higgs Field equipped with a tetrahedron lattice structure with variable Planck length around black holes..
3: Quantum Gravity = Dual Push gravity= Attraction (Higgs-Casimir opposing Graviton push).
4: The Big Bang is a Splitting dark matter Big Bang Black Hole (BBBH), splitting into smaller Primordial Big Bang Spinters (PBBS) forming the Fractalic Lyman Alpha forest and evaporating partly into a zero mass energetic oscillating Higgs particle based Higgs field.
5: Dual PBBSs hotspots, produce central plasma concentration in electric Herbig Haro systems as a base for star formation in open star clusters as a start for Spiral Galaxies.
6: Spiral Galaxies will keep both Primordial Dark Matter Black Holes as Galaxy Anchor Black Holes (GABHs) at long distance.
7: After Galaxy Merging, these GABHs are the origin of Galaxy- and Magnetic field complexity and distant dwarf galaxies .
8: Black Holes produce Plasma direct out of the Higgs field because two rigid massless but mutable Higgs particles are convertible into symmetric electron and positron (or even dual quark-) propellers (by BH horizon fluctuations).
9: The chirality of the (spiralling) vacuum lattice is the origin our material universe. (propeller shaped positrons merge preferentially first with gluons to form (u) Quarks to form Hydrogen with electrons.
10: The first Supernovas produce medium sized Black Holes as the base for secondary Herbig Haro systems and open star clusters.
11: ALL Dark Matter Black Holes are supposed to be CHARGE SEPARATORS with internal positive charge and an external globular shell of negative charged Quark electron plasma.
12: The lightspeed is related to gravity fields like the earth with long extinction distances to adapt with the solar gravity field.
13. Quantum FFF Theory states that the raspberry shaped multiverse is symmetric and instant entangled by so called WORMHOLE BRIDGES (ER=EPR) down to the smallest quantum level between different raspberry berries=universes. Nonlocal entangled down to each quantum and even living or dying CATS in BOXES.
14 Large Primordial Big Bang Spinters (PBBS) are responsible for the creation of the Lyman Alpha forest structure and first spiral galaxy forming of the universe, but they seem to be also responsible for the implosion of the universe at the end in the form of Galaxy Anchor Black Holes (GABHs) located mainly outside galaxies. see: (Quasisoft Chandra sources)
15, As a consequence of 12-14, Time is always local time in combination with cyclic period time of the multiverse.
16, If our material universes has a chiral left handed oscillating Higgs field, then our material Right Handed DNA helix molecule could be explained.

However it also suggests that in our opposing symmetric ANTI-MATERIAL entangled neighbor universe the DNA helix should have a Left handed spiral.
According to Max Tegmark: in an entangled multiverse we may ask: is there COPY PERSON over there, who is reading the same lines as I do?
If this COPY person is indeed living over there, then even our consciousness should be shared in a sort of entangled DEMOCRATIC form.
Then we are not alone with our thoughts and doubts, see:
Democratic Free Will in the instant Entangled Multiverse.

17, All particles and waves should be entangled with anti-copy particles and Higgs field based waves in our opposing anti universes.
Then wave particle duality and the so called one particle double slit interference could be explained by accepting that real shaped 3D photon particles only exist in their function of Gluons. However Gluons should transfer form information to the oscillating luminiferous Higgs field ether, within about 1 cm distance from a fermion.(see: Babock and Bergman anomaly,1964)
18, Different elementary particles have different qualities by their different complex stringy shape. Leptons and Quarks have a propeller shape with left or right handed pitch creating charge difference. Gluons, Photons and Neutrino particles have no pitch.
20 All particles can be converted to one or more Higgs particles, Lepto Quarks are also compound particles with an exeption for the electron and positron..
21 All particles are assumed to be present as Higgs particles already in the cyclic Big Bang Black Hole (BBBH) after the former multiverse has collapsed into a Big Crunch Black Hole.
22, We seem to live inside one material universal bubble of an instant entangled (Charge Parity Time) CPT symmetric raspberry bubble multiverse. Each quantum jump or wave function collapse or human choice is guided by this instant entanglement at long distant.

If our material universes has a chiral oscillating Higgs field, then our material Right Handed DNA helix molecule could be explained.
However it also suggests that in our opposing ANTI-MATERIAL multiverse neighbour universe the DNA helix should have a LEFT HANDED spiral.
According to Max Tegmark: in an entangled multiverse we may ask: is there COPY PERSON over there, who is reading the same lines as I do?
If this COPY person is indeed living over there, then even our consciousness should be shared in a sort of DEMOCRATIC form,
Then we are not alone with our thoughts and doubts, see:
Democratic Free Will in the instant Entangled Multiverse.
Conclusion, Quantum FFF Theory is based on Non-Local Realism. The CP(T)-symmetrical raspberry shaped multiverse (with 8-12 berries) is non local instant connected down to each quantum. Each quantum can be visualised as a deformed Axion-Higgs ring able to convert into other shapes and able to form compound structures like quarks, muons or even Tau leptons. So realism (cogwheels) connected by symmetric non-local instant entanglement.
vixra.org/pdf/1401.0071v2.pdf

It was Abdus Salam who proposed that quarks and leptons should have a sub-quantum level structure, and that they are compound hardrock particles with a specific non-zero sized form. Jean Paul Vigier postulated that quarks and leptons are "pushed around" by an energetic sea of vacuum particles.
David Bohm suggested in contrast with The "Copenhagen interpretation", that reality is not created by the eye of the human observer, and second: elementary particles should be "guided by a pilot wave".
John Bell argued that the motion of mass related to the surrounding vacuum reference frame, should originate real
"Lorentz-transformations", and also real relativistic measurable contraction.
Richard Feynman postulated the idea of an all pervading energetic quantum vacuum. He rejected it, because it should originate resistance for every mass in motion, relative to the reference frame of the quantum vacuum.
However, I postulate the strange and counter intuitive possibility, that this resistance for mass in motion, can be compensated, if we combine the ideas of Vigier, Bell, Bohm and Salam, and a new dual universal Bohmian "pilot wave", which is interpreted as the EPR correlation (or Big Bang entanglement) between individual elementary anti-mirror particles, living in dual universes.

Leo Vuyk (Quantum FFF Theory Posters)
A rigid string theory.
Rigid strings able to mutate and merge by interlocking into (lepto) Quarks. The architecture of the quantum world based on a massless single Higgs torus able to mutate into electron or positron or photon.
So photon radiation is only the transfer by oscillating encoded Higgs particles living inside the space time lattice.
Fermions have to work hard against bouncing Higgses to pass this lattice and mutate Higgses into real rigid photons by Higgs collision. After about 1 cm this rigid photon will click back as ancoded Higgs field particle which seems to be the base for general Relativity including an extra LASOF ether effect.

### An experimental verified theory for free will. as a base for a Theory Of Everything (TOE)

It is well known that not only elements of the standard model combined with Quantum mechanics and relativity rules are needed to get a shimmer of a TOE.
However a TOE should be combined with an experimental verified model for free will.

The Quantum FFF model (Function Follows Form) seems to point to a solution of a raspberry shaped instant entangled Charge Parity  symmetric multiverse.

See two articles:
Democratic Free Will in the instant Entangled Multiverse.

And:
New physics elements in the Quantum Function Follows Form Model.

Quantum FFF Theory is also published in the form of POSTERS at the Flickr site: http://www.flickr.com/photos/93308747@N05/

Numbered listing of Vixra essays by Leo Vuyk.

## Thursday, October 20, 2016

### Curious tilt of the Sun traced to undiscovered planet? or by a remnant HH system

Or the Herbig Haro remnant black hole system as Quantum FFF Theory predict.
see: my former blog:  http://bigbang-entanglement.blogspot.nl/2016/10/irs43-new-physics-by-4x-herbig-haro.html
See first: https://www.sciencedaily.com/releases/2016/10/161019173023.htm

All of the planets orbit in a flat plane with respect to the Sun, roughly within a couple degrees of each other.
That plane, however, rotates at a six-degree tilt with respect to the Sun giving the appearance that the Sun itself is cocked off at an angle.
Until now, no one had found a compelling explanation to produce such an effect.
"It's such a deep-rooted mystery and so difficult to explain that people just don't talk about it," says Brown,
the Richard and Barbara Rosenberg Professor of Planetary Astronomy.

Other indications for remnant Herbig Haro Black holes around the Sun.

## Sunday, October 16, 2016

### IRS43: New Physics by 4x Herbig Haro Black Holes ( HH knots).

IRS43: New Physics by 4x Herbig Haro Black Holes.

Supported by the surprising third gas/planetary disc around binary IRS 43.   According to Quantum FFF Theory, all H-H systems are based on new physics tandem dark matter black holes ( HH knots) with star formation in between.  A double H-H system should be based on 4x black holes, which also seem to be able to form a third gas disc in between.
It is assumed that the space(time) between all H-H knots is polarized with a quantum gravity- and electric  effect on all matter in between the knots, resulting in a sort of gas-dust sweeping pressure cooker effect  for star formation.
For background model info: http://vixra.org/pdf/1609.0407v1.pdf
and: http://vixra.org/author/leo_vuyk

NEWS:
Curious tilt of the Sun traced to undiscovered planet?
Or the Herbig Haro remnant black hole system as Quantum FFF Theory predict.
https://www.sciencedaily.com/releases/2016/10/161019173023.htm

All of the planets orbit in a flat plane with respect to the Sun, roughly within a couple degrees of each other.
That plane, however, rotates at a six-degree tilt with respect to the Sun giving the appearance that the Sun itself is cocked off at an angle.
Until now, no one had found a compelling explanation to produce such an effect.
"It's such a deep-rooted mystery and so difficult to explain that people just don't talk about it," says Brown,
the Richard and Barbara Rosenberg Professor of Planetary Astronomy.

Upper panel: the region of the IRS 54 outflow as well as the near-infrared Herbig-Haro objects from Spitzer observations with 3.6 $\mu$m in blue, 4.5 $\mu$m in green and 8.0$~\mu$m in red.
In addition we show the suggested direction of the precessing IRS 54 outflow (dotted line) and the direction to IRS 43 and IRS 44 (dashed arrow). Lower panels: images of the near-infrared Herbig-Haro knots from 3.6-8.0 $\mu$m (IRAC1-4) and 24 $\mu$m (MIPS1).
The location of the Herbig-Haro knots suggested by Grosso et al. (2001) to be associated with the IRS 54 outflow are shown with the crosses, whereas the knots proposed to be associated with the IRS 43 outflow are shown with plus-signs in the first panel.

(Fig. B.1) show the knots clearly along a line pointing directly back toward the group of objects where IRS 43 and another embedded protostar, IRS 44, is located. As also pointed out by Grosso et al., the association of the Herbig-Haro knots to IRS 43 is not unique: the nearby protostar, IRS 54 (see also Sect. B.5), also drives a large-scale, precessing outflow clearly seen in the Spitzer images in Fig. B.1.

IRS 43 compared with Cat's eye and Twin jet PN

How different sized new physics black holes seem to behave as mergers or in dual repelling  twin action such as Herbig Haro systems

Black hole pairing and repelling by equaly sized BHs.

Stars and Galaxies show equal spiral systems with 2x or 4x or more arms.

Quantum FFF Theory principles.
Quantum FFF Theory is based on Non-Local Realism and rigid stringy Axion-Higgs particles able to mutate into all other particles.
The CP(T)-symmetrical raspberry shaped multiverse (with 8-12 berries) is non local instant connected down to each quantum. Each quantum can be visualised as a deformed Higgs ring able to convert into other shapes and able to form compound structures like quarks, muons or even Tau leptons. So realism (cogwheels) connected by symmetric non-local instant entanglement.

The basic elements:
1: Black holes (BHs) are the same as Dark Matter, they all consume photons, even gravitons and the Axion-Higgs field, ( for short: Higgs field) but REPEL Fermions due to their propeller shape polarisation away from the BH. They produce electric charged plasma.
2: Dark Energy is the oscillating ( Casimir) energy of the massless Axion-Higgs Field equipped with a tetrahedron lattice structure with variable Planck length around black holes..
3: Quantum Gravity = Dual Push gravity= Attraction (Higgs-Casimir opposing Graviton push).
4: The Big Bang is a Splitting dark matter Big Bang Black Hole (BBBH), splitting into smaller Primordial Big Bang Spinters (PBBS) forming the Fractalic Lyman Alpha forest and evaporating partly into a zero mass energetic oscillating Higgs particle based Higgs field.
5: Dual PBBSs hotspots, produce central plasma concentration in electric Herbig Haro systems as a base for star formation in open star clusters as a start for Spiral Galaxies.
6: Spiral Galaxies will keep both Primordial Dark Matter Black Holes as Galaxy Anchor Black Holes (GABHs) at long distance.
7: After Galaxy Merging, these GABHs are the origin of Galaxy- and Magnetic field complexity and distant dwarf galaxies .
8: Black Holes produce Plasma direct out of the Higgs field because two rigid massless but mutable Higgs particles are convertible into symmetric electron and positron (or even dual quark-) propellers (by BH horizon fluctuations).
9: The chirality of the (spiralling) vacuum lattice is the origin our material universe. (propeller shaped positrons merge preferentially first with gluons to form (u) Quarks to form Hydrogen with electrons.
10: The first Supernovas produce medium sized Black Holes as the base for secondary Herbig Haro systems and open star clusters.
11: ALL Dark Matter Black Holes are supposed to be CHARGE SEPARATORS with internal positive charge and an external globular shell of negative charged Quark electron plasma.
12: The lightspeed is related to gravity fields like the earth with long extinction distances to adapt with the solar gravity field.
13. Quantum FFF Theory states that the raspberry shaped multiverse is symmetric and instant entangled by so called WORMHOLE BRIDGES (ER=EPR) down to the smallest quantum level between different raspberry berries=universes. Nonlocal entangled down to each quantum and even living or dying CATS in BOXES.
14 Large Primordial Big Bang Spinters (PBBS) are responsible for the creation of the Lyman Alpha forest structure and first spiral galaxy forming of the universe, but they seem to be also responsible for the implosion of the universe at the end in the form of Galaxy Anchor Black Holes (GABHs) located mainly outside galaxies. see: (Quasisoft Chandra sources)
15, As a consequence of 12-14, Time is always local time in combination with cyclic period time of the multiverse.
16, If our material universes has a chiral left handed oscillating Higgs field, then our material Right Handed DNA helix molecule could be explained.

However it also suggests that in our opposing symmetric ANTI-MATERIAL entangled neighbor universe the DNA helix should have a Left handed spiral.
According to Max Tegmark: in an entangled multiverse we may ask: is there COPY PERSON over there, who is reading the same lines as I do?
If this COPY person is indeed living over there, then even our consciousness should be shared in a sort of entangled DEMOCRATIC form.
Then we are not alone with our thoughts and doubts, see:
Democratic Free Will in the instant Entangled Multiverse.

17, All particles and waves should be entangled with anti-copy particles and Higgs field based waves in our opposing anti universes.
Then wave particle duality and the so called one particle double slit interference could be explained by accepting that real shaped 3D photon particles only exist in their function of Gluons. However Gluons should transfer form information to the oscillating luminiferous Higgs field ether, within about 1 cm distance from a fermion.(see: Babock and Bergman anomaly,1964)
18, Different elementary particles have different qualities by their different complex stringy shape. Leptons and Quarks have a propeller shape with left or right handed pitch creating charge difference. Gluons, Photons and Neutrino particles have no pitch.
20 All particles can be converted to one or more Higgs particles, Lepto Quarks are also compound particles with an exeption for the electron and positron..
21 All particles are assumed to be present as Higgs particles already in the cyclic Big Bang Black Hole (BBBH) after the former multiverse has collapsed into a Big Crunch Black Hole.
22, We seem to live inside one material universal bubble of an instant entangled (Charge Parity Time) CPT symmetric raspberry bubble multiverse. Each quantum jump or wave function collapse or human choice is guided by this instant entanglement at long distant.

If our material universes has a chiral oscillating Higgs field, then our material Right Handed DNA helix molecule could be explained.
However it also suggests that in our opposing ANTI-MATERIAL multiverse neighbour universe the DNA helix should have a LEFT HANDED spiral.
According to Max Tegmark: in an entangled multiverse we may ask: is there COPY PERSON over there, who is reading the same lines as I do?
If this COPY person is indeed living over there, then even our consciousness should be shared in a sort of DEMOCRATIC form,
Then we are not alone with our thoughts and doubts, see:
Democratic Free Will in the instant Entangled Multiverse.
Conclusion, Quantum FFF Theory is based on Non-Local Realism. The CP(T)-symmetrical raspberry shaped multiverse (with 8-12 berries) is non local instant connected down to each quantum. Each quantum can be visualised as a deformed Axion-Higgs ring able to convert into other shapes and able to form compound structures like quarks, muons or even Tau leptons. So realism (cogwheels) connected by symmetric non-local instant entanglement.
vixra.org/pdf/1401.0071v2.pdf

It was Abdus Salam who proposed that quarks and leptons should have a sub-quantum level structure, and that they are compound hardrock particles with a specific non-zero sized form. Jean Paul Vigier postulated that quarks and leptons are "pushed around" by an energetic sea of vacuum particles.
David Bohm suggested in contrast with The "Copenhagen interpretation", that reality is not created by the eye of the human observer, and second: elementary particles should be "guided by a pilot wave".
John Bell argued that the motion of mass related to the surrounding vacuum reference frame, should originate real
"Lorentz-transformations", and also real relativistic measurable contraction.
Richard Feynman postulated the idea of an all pervading energetic quantum vacuum. He rejected it, because it should originate resistance for every mass in motion, relative to the reference frame of the quantum vacuum.
However, I postulate the strange and counter intuitive possibility, that this resistance for mass in motion, can be compensated, if we combine the ideas of Vigier, Bell, Bohm and Salam, and a new dual universal Bohmian "pilot wave", which is interpreted as the EPR correlation (or Big Bang entanglement) between individual elementary anti-mirror particles, living in dual universes.

Leo Vuyk (Quantum FFF Theory Posters)
A rigid string theory.
Rigid strings able to mutate and merge by interlocking into (lepto) Quarks. The architecture of the quantum world based on a massless single Higgs torus able to mutate into electron or positron or photon.
So photon radiation is only the transfer by oscillating encoded Higgs particles living inside the space time lattice.
Fermions have to work hard against bouncing Higgses to pass this lattice and mutate Higgses into real rigid photons by Higgs collision. After about 1 cm this rigid photon will click back as ancoded Higgs field particle which seems to be the base for general Relativity including an extra LASOF ether effect.


text:
# I Using determinant to find constraints on equation

Tags:
1. Jan 15, 2017

### TheDemx27

Basically I don't know how to get to the constraints from the system of equations. In class we used det to find constraints for homogenous equations, but we didn't go over this situation. Someone spell it out for me?

2. Jan 15, 2017

### mathman

The determinant is $sin^2(\theta )(1+cos^2(\theta ))$.

3. Jan 15, 2017

### TheDemx27

How did you get that?
$det(A)=cos^2(\theta )sin^2(\theta )+sin^4(\theta )$ Which then simplifies into what they got: $=\frac{1}{2}\ (1-cos(\theta ))=sin^2(\theta )$

My question is how they proceed after that anyways.

4. Jan 15, 2017

### Stephen Tashi

The solution for a particular unknown in a system of linear equations can be expressed as a ratio of two determinants. In your example, the determinant of the denominator is $\sin^2(q)$. The numerator is the determinant of a matrix formed by using $C_1, C_2$ in place of some entries in the coefficient matrix, depending on which unknown you are solving for. See if the standard method of solving equations in that manner ( e.g. https://www.cliffsnotes.com/study-g...tions-using-determinants-with-three-variables ) derives the equations the article ends up with.

$\begin{pmatrix} \cos(\theta) \sin(\theta) & \sin^2(\theta) \\ -\sin^2(\theta) & \cos(\theta)\sin(\theta) \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} =\begin{pmatrix} C_1 \\ C_2 \end{pmatrix}$

Solve for $X,Y$.

5. Jan 16, 2017

my mistake


text:
1. ## Really hard questions...

1. XY and YZ are diameters of two tangent circles and XY = 10 cm. Find YZ if the area of hte shaded region equals the area of hte unshaded region.

The answer is: 24.14, but how and why?

2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

Notes: These drawings are not EXACT as the real one because I used paint, however they really close.

The answer is 4 cm and 8 cm, but how and why...?

These questions are really hard. Nobody in my class were able to answer them.

2. ## Suggested soln

ok.. here we go...
let's say the dist YZ is x cm, then the diameter of the larger circle is (10+x) cm. And the area is pi [(10+x)/2]^2. Similarly, the area of the smaller circle can be found ie pi(x/2)^2. We can then find an expression for the shaded area and equate that to the area of the smaller circle. You should get a wuadratic equation which can be reduced to
x^2 - 20x -100 = 0.
Solving this would give you the answer as required.

As for the second question, I think you should work on it alonmg similar lines as per question 1.

Hope it helps. Let me know if u need further clarifications.

musMD, Singapore

3. Originally Posted by AlphaRock
[B]...

2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

...
to #2:

I've modified your sketch a little bit.

1. Let x denote the length of the diameter of the upper circle then (12-x) is the length of the diameter of the lower circle.

2. The shaded region is the difference of the large circle and the two small circles:

$A=\pi \cdot \left(\frac{12}2 \right)^2 - \pi \cdot \left(\frac{x}2 \right)^2 - \pi \cdot \left(\frac{12-x}2 \right)^2 = \frac49 \cdot \pi \cdot \left(\frac{12}2 \right)^2$

3. Expand the brackets, collect like terms. You'll get a quadratic equation:

$-\frac12x^2+6x-16=0$

Solve for x.


text: Given a right triangle with an acute angle $\theta$, the side of the triangle that is adjacent to $\theta$ is the side of the triangle that is also a side (http://planetmath.org/Angle) of $\theta$ and is not the hypotenuse.