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What’s Happening in the Mathematical Sciences, Volume 6

Available Formats:
Softcover ISBN: 978-0-8218-3585-2
Product Code: HAPPENING/6
List Price: $28.00 MAA Member Price:$25.20
AMS Member Price: $22.40 Click above image for expanded view What’s Happening in the Mathematical Sciences, Volume 6 Available Formats:  Softcover ISBN: 978-0-8218-3585-2 Product Code: HAPPENING/6  List Price:$28.00 MAA Member Price: $25.20 AMS Member Price:$22.40
• Book Details

What's Happening in the Mathematical Sciences
Volume: 62006; 122 pp
MSC: Primary 00;

The AMS series What's Happening in the Mathematical Sciences distills the amazingly rich brew of current research in mathematics down to a few choice samples. This volume leads off with an update on the Poincaré Conjecture, a hundred-year-old problem that has apparently been solved by Grigory Perelman of St. Petersburg, Russia. So what did topologists do when the oldest and most famous problem about closed manifolds was vanquished? As the second chapter describes, they confronted a suite of problems concerning the “ends” of open manifolds… and solved those, too.

Not to be outdone, number theorists accomplished several unexpected feats in the first five years of the new century, from computing a trillion digits of pi to finding arbitrarily long equally-spaced sequences of prime numbers. Undergraduates made key discoveries, as explained in the chapters on Venn diagrams and primality testing. In applied mathematics, the Navier-Stokes equations of fluid mechanics continued to stir up interest. One team proved new theorems about the long-term evolution of vortices, while others explored the surprising ways that insects use vortices to move around. The random jittering of Brownian motion became a little less mysterious. Finally, an old and trusted algorithm of computer science had its trustworthiness explained in a novel way.

Barry Cipra explains these new developments in his wry and witty style, familiar to readers of Volumes 1–5, and is joined in this volume by Dana Mackenzie. Volume 6 of What's Happening will convey to all readers—from mathematical novices to experts—the beauty and wonder that is mathematics.

Click here to listen to interviews with authors Barry Cipra and Dana Mackenzie.

General mathematical audience.

• Request Review Copy
Volume: 62006; 122 pp
MSC: Primary 00;

The AMS series What's Happening in the Mathematical Sciences distills the amazingly rich brew of current research in mathematics down to a few choice samples. This volume leads off with an update on the Poincaré Conjecture, a hundred-year-old problem that has apparently been solved by Grigory Perelman of St. Petersburg, Russia. So what did topologists do when the oldest and most famous problem about closed manifolds was vanquished? As the second chapter describes, they confronted a suite of problems concerning the “ends” of open manifolds… and solved those, too.

Not to be outdone, number theorists accomplished several unexpected feats in the first five years of the new century, from computing a trillion digits of pi to finding arbitrarily long equally-spaced sequences of prime numbers. Undergraduates made key discoveries, as explained in the chapters on Venn diagrams and primality testing. In applied mathematics, the Navier-Stokes equations of fluid mechanics continued to stir up interest. One team proved new theorems about the long-term evolution of vortices, while others explored the surprising ways that insects use vortices to move around. The random jittering of Brownian motion became a little less mysterious. Finally, an old and trusted algorithm of computer science had its trustworthiness explained in a novel way.

Barry Cipra explains these new developments in his wry and witty style, familiar to readers of Volumes 1–5, and is joined in this volume by Dana Mackenzie. Volume 6 of What's Happening will convey to all readers—from mathematical novices to experts—the beauty and wonder that is mathematics.

Click here to listen to interviews with authors Barry Cipra and Dana Mackenzie.


text:
# How to draw a proper decision tree

I need to draw a decision tree for my LaTeX document (with TikZ/forest or any other package). I searched around for a while to find something similar, but found none. The decision tree produced should be like the one shown in the figure bellow.

Any help/hint would be very much appreciated.

This is what I managed to do so far, of course the shape of the tree does not have to be exactly the same as the one in picture.

\documentclass[margin=10pt]{standalone}
\usepackage{tikz,forest}
\usetikzlibrary{arrows.meta}

\begin{document}
\tikzset{
decision/.style={diamond, minimum height=10pt, minimum width=10pt, inner sep=1pt},
chance/.style={circle, minimum width=10pt, draw=blue, fill=none, thick, inner sep=0pt},
}
\begin{forest}
label L/.style={
edge label={node[midway,left,font=\scriptsize]{#1}}
},
label R/.style={
edge label={node[midway,right,font=\scriptsize]{#1}}
},
for tree={
child anchor=north,
for descendants={
{edge=->}
}
},
[$x_2$, decision, draw
[$x_1$, decision, draw, label L=N,
[1, rectangle, draw, label L=N, tier=bottom]
[0, rectangle, draw, label R=Y, tier=bottom]
]
[$x_3$, decision, draw, label R=Y,
[$x_1$, decision, draw, label L=N,
[1, rectangle, draw, label L=N, tier=bottom]
[0, rectangle, draw, label R=Y, tier=bottom]
]
[0, rectangle, draw, label R=Y, tier=bottom]
]
]
\end{forest}

\end{document}

• Please show us the code for some basic forest or tikz-tree solution and we will be happy to help you with any issue, you find. Like this, it is just a "do it for me" post. – LaRiFaRi Jan 27 '16 at 11:24
• I added the code of what I have so far. – Trex Jan 27 '16 at 12:26

Here's a solution with forest.

## Code

\documentclass[margin=10pt]{standalone}
\usepackage{tikz,forest}
\usetikzlibrary{arrows.meta}

\forestset{
.style={
for tree={
base=bottom,
child anchor=north,
align=center,
s sep+=1cm,
straight edge/.style={
edge path={\noexpand\path[\forestoption{edge},thick,-{Latex}]
(!u.parent anchor) -- (.child anchor);}
},
if n children={0}
{tier=word, draw, thick, rectangle}
{draw, diamond, thick, aspect=2},
if n=1{%
edge path={\noexpand\path[\forestoption{edge},thick,-{Latex}]
(!u.parent anchor) -| (.child anchor) node[pos=.2, above] {Y};}
}{
edge path={\noexpand\path[\forestoption{edge},thick,-{Latex}]
(!u.parent anchor) -| (.child anchor) node[pos=.2, above] {N};}
}
}
}
}

\begin{document}
\begin{forest}
[$x_2$, tikz={\draw[{Latex}-, thick] (.north) --++ (0,1);}
[$x_1$
[1]
[0]
]
[$x_3$
[$x_1$
[1]
[0]
]
[0]
]
]
\end{forest}
\end{document}

• looks very nice. How to add an edge to the root node (as shown in the original figure I posted) ? – Trex Jan 27 '16 at 14:28
• @Trex I thought that was not to be placed. Where is the root node? Is it a single node above? – Alenanno Jan 27 '16 at 14:30
• The root node is the top node, node $x_2$ in your sample. There should be an edge pointing downwards to it. – Trex Jan 27 '16 at 14:34
• @Trex An edge without another node? I mean, just the edge? – Alenanno Jan 27 '16 at 14:38
• yeah, just like in the original figure – Trex Jan 27 '16 at 14:40
\documentclass{standalone}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english,italian]{babel}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows,fit,calc,positioning}
\tikzset{box/.style={draw, diamond, thick, text centered, minimum height=0.5cm, minimum width=1cm}}
\tikzset{line/.style={draw, thick, -latex'}}
\begin{document}
\begin{tikzpicture}[auto]
\node [box]                                    (x3)      {x\ped{3}};
\node [box, below=0.5cm of x3, xshift=-3cm]    (x1sx)    {x\ped{1}};
\node [box, below=0.5cm of x3, xshift=3cm]     (x2dx)    {x\ped{2}};
\node [box, below=0.5cm of x1sx, xshift=-1cm]  (x2sx)    {x\ped{2}};
\node [box, below=0.5cm of x2sx, xshift=1cm]   (A2sx)    {A\ped{2}};
\node [box, below=0.5cm of x2sx, xshift=-1cm]  (A1sx)    {A\ped{1}};
\node [box, right=1cm of A2sx]                 (A3sx)    {A\ped{3}};
%
\node [box, below=0.5cm of x2dx, xshift=1cm]   (x1dx)    {x\ped{1}};
\node [box, below=0.5cm of x1dx, xshift=-1cm]  (A2dx)    {A\ped{2}};
\node [box, below=0.5cm of x1dx, xshift=1cm]   (A3dx)    {A\ped{3}};
\node [box, left=0.5cm of A2dx]                (A1dx)    {A\ped{1}};
%
\path [line] (x3) -|         (x2dx);
\path [line] (x3) -|         (x1sx);
\path [line] (x2dx) -|       (x1dx);
\path [line] (x2dx) -|       (A1dx);
\path [line] (x1dx) -|       (A2dx);
\path [line] (x1dx) -|       (A3dx);
\path [line] (x1sx) -|       (x2sx);
\path [line] (x1sx) -|       (A3sx);
\path [line] (x2sx) -|       (A1sx);
\path [line] (x2sx) -|       (A2sx);
\end{tikzpicture}
\end{document}

• Thanks, it looks nice. However, is it possible to do it with forest? Furthermore, how can I add an edge to the root node, as shown in the original picture? – Trex Jan 27 '16 at 14:05
• I'm trying to figure out how to add Y and N. I'll edit my message if I'm able to come up with a solution. – Marco Jan 27 '16 at 14:11
• I had to remove italian from babel and add \newcommand{\ped}[1]{$_{#1}$} to the preamble for this to compile. – AstroFloyd Mar 8 at 7:39

format: plain


text:
# Two formulas for complex inner product: Which formula is true?

I am a bit confused regarding the correct formula for the complex inner product of two complex vectors. The textbook I learnt from defines the complex inner product of the vectors: $u$ and $v$ as follows: $$\left ( cu,v \right )=\bar{c}\left ( u,v \right )$$ and $$\left ( u,cv \right )=c\left ( u,v \right )$$ where $c$ is a scalar.

However, in a different textbook, I found the following convention: $$\left ( cu,v \right )=c\left ( u,v \right )$$ and $$\left ( u,cv \right )=\bar{c}\left ( u,v \right )$$

Which one of the above formulas is the one I should use?

Also, for the complex dot product: In some textbooks, I found: $$u\cdot v=\bar{u_{1}}v_{1}+\bar{u_{2}}v_{2}+\cdots+\bar{u_{n}}v_{n}.$$ In other textbooks, I found the following: $$u\cdot v= u_{1}\bar{v_{1}}+u_{2}\bar{v_{2}}+\cdots+u_{n}\bar{v_{n}}$$

• Which one I should as the definition of the complex dot product? Also, which one is equivalent to $u^{*}v$?
-
Either one works; it depends on your context. Mathematicians prefer the second one (linear in the first coordinate, conjugate linear in the second). I believe physicists prefer the former (linear in the second coordinate, conjugate linear in the first coordinate). So long as you adapt formulas appropriately, they give you the same results. – Arturo Magidin Feb 24 '12 at 20:35
If you write your vectors as columns, $u=(u_1,\ldots,u_n)^t$ and $v = (v_1,\ldots,v_n)^t$, then $u^* = (\overline{u_1},\ldots,\overline{u_n})$, so $u^*v$ is given by your first definition of "complex dot product." – Arturo Magidin Feb 24 '12 at 20:37

"True" is not the right word to use here. The two conventions are interchangeable, since given any inner product of the first kind defining $(u, v)_{\text{new}} = (v, u)_{\text{old}}$ gives an inner product of the second kind. So it doesn't matter which one you use, really, as long as you are consistent.
Nevertheless, I would like to argue that the convention where the inner product is conjugate-linear in the first variable is the "better" one to use. One bit of evidence comes from the fact that $\dim \text{Hom}(V, W)$ naturally appears in the representation theory of finite groups, and it gives an inner product on characters which is conjugate-linear in the first variable. The discussion in the beginning of Baez's Higher-Dimensional Algebra II: 2-Hilbert Spaces is relevant.
Unfortunately, there are two different conventions floating around out there. Either one is correct, but you need to be very careful that you define everything which depends on the inner product in a manner consistent with the particular definition of inner product you are using. For example, if you use your first definition of inner product then $$u\cdot v=\sum_{i=1}^n\langle u_ie_i,v_ie_i\rangle = \sum_{i=1}^n \bar{u_i}\cdot v_i$$ where $e_1,\ldots,e_n$ is the standard basis, while if you use the second then $$u\cdot v=\sum_{i=1}^n\langle u_ie_i,v_ie_i\rangle = \sum_{i=1}^n u_i\cdot \bar{v_i}$$ and similar modifications have to be made to other definitions and theorems. The moral: be sure to check what definition your professor and textbook use (hope that they are the same!) and if you every become a brilliant mathematician and invent some new math, insist on just one convention.


text:
That is what I envision. Then you have the vehicle I will build. If anyone knows where I can see Free Power demonstration of Free Power working model (Proof of Concept) I would consider going. Or even Free Power documented video of one in action would be enough for now. Burp-Professor Free Power Gaseous and Prof. Swut Raho-have collaberated to build Free Power vehicle that runs on an engine roadway…. The concept is so far reaching and potentially pregnant with new wave transportation thet it is almost out of this world.. Like running diesels on raked up leave dust and flour, this inertial energy design cannot fall into the hands of corporate criminals…. Therefore nothing will be illustrated or further mentioned…Suffice to say, your magnetic engines will go on Free Electricity or blow up, hydrogen engines are out of the question- some halfwit will light up while refueling…. America does not deserve the edge anymore, so look to Europe, particuliarly the scots to move transportation into the Free Electricity century…
The “energy ” quoted in magnetization is the joules of energy required in terms of volts and amps to drive the magnetizing coil. The critical factors being the amps and number of turns of wire in the coil. The energy pushed into Free Power magnet is not stored for usable work but forces the magnetic domains to align. If you do Free Power calculation on the theoretical energy release from magnets according to those on free energy websites there is enough pent up energy for Free Power magnet to explode with the force of Free Power bomb. And that is never going to happen. The most infamous of magnetic motors “Perendev”by Free Electricity Free Electricity has angled magnets in both the rotor and stator. It doesn’t work. Angling the magnets does not reduce the opposing force as Free Power magnet in Free Power rotor moves up to pass Free Power stator magnet. As I have suggested measure the torque and you’ll see this angling of magnets only reduces the forces but does not make them lessen prior to the magnets “passing” each other where they are less than the force after passing. Free Energy’t take my word for it, measure it. Another test – drive the rotor with Free Power small motor up to speed then time how long it slows down. Then do the same test in reverse. It will take the same time to slow down. Any differences will be due to experimental error. Free Electricity, i forgot about the mags loseing their power.
I am currently designing my own magnet motor. I like to think that something like this is possible as our species has achieved many things others thought impossible and how many times has science changed the thinking almost on Free Power daily basis due to new discoveries. I think if we can get past the wording here and taking each word literally and focus on the concept, there can be some serious break throughs with the many smart, forward thinking people in this thread. Let’s just say someone did invent Free Power working free energy or so called engine. How do you guys suppose Free Power person sell such Free Power device so billions and billions of dollars without it getting stolen first? Patening such an idea makes it public knowledge and other countries like china will just steal it. Such Free Power device effects the whole world. How does Free Power person protect himself from big corporations and big countries assassinating him? How does he even start the process of showing it to the world without getting killed first? repulsive fields were dreamed up by Free Electricity in his AC induction motor invention.
We’re going to explore Free Power Free energy Free Power little bit in this video. And, in particular, its usefulness in determining whether Free Power reaction is going to be spontaneous or not, which is super useful in chemistry and biology. And, it was defined by Free Power Free Energy Free Power. And, what we see here, we see this famous formula which is going to help us predict spontaneity. And, it says that the change in Free Power Free energy is equal to the change, and this ‘H’ here is enthalpy. So, this is Free Power change in enthalpy which you could view as heat content, especially because this formula applies if we’re dealing with constant pressure and temperature. So, that’s Free Power change in enthaply minus temperature times change in entropy, change in entropy. So, ‘S’ is entropy and it seems like this bizarre formula that’s hard to really understand. But, as we’ll see, it makes Free Power lot of intuitive sense. Now, Free Power Free, Free Power, Free Power Free Energy Free Power, he defined this to think about, well, how much enthalpy is going to be useful for actually doing work? How much is free to do useful things? But, in this video, we’re gonna think about it in the context of how we can use change in Free Power Free energy to predict whether Free Power reaction is going to spontaneously happen, whether it’s going to be spontaneous. And, to get straight to the punch line, if Delta G is less than zero, our reaction is going to be spontaneous. It’s going to be spontaneous. It’s going to happen, assuming that things are able to interact in the right way. It’s going to be spontaneous. Now, let’s think Free Power little bit about why that makes sense. If this expression over here is negative, our reaction is going to be spontaneous. So, let’s think about all of the different scenarios. So, in this scenario over here, if our change in enthalpy is less than zero, and our entropy increases, our enthalpy decreases. So, this means we’re going to release, we’re going to release energy here. We’re gonna release enthalpy. And, you could think about this as, so let’s see, we’re gonna release energy. So, release. I’ll just draw it. This is Free Power release of enthalpy over here.
My older brother explained that in high school physics, they learned that magnetism is not energy at all. Never was, never will be. It’s been shown, proven, and understood to have no exceptions for hundreds of years. Something that O. U. should learn but refuses to. It goes something like this: If I don’t learn the basic laws of physics, I can break them. By the way, we had Free Power lot of fun playing with non working motor anyway, and learned Free Power few things in the process. My brother went on to get his PHD in physics and wound up specializing in magnetism. He designed many of the disk drive plates and electronics in the early (DOS) computers. bnjroo Harvey1 Thanks for the reply! I’m afraid there is an endless list of swindlers and suckers out there. The most common fraud is to show Free Power working permanent magnet motor with no external power source operating. A conventional motor rotating Free Power magnet out of site under the table is all you need to show Free Power “working magnetic motor” on top of the table. How could I know this? Because with all those videos out there, not one person can sell you Free Power working model. Also, not one of these scammers can ever let anyone not related to his scam operate the motor without the scammer hovering around. The believers are victims of something called “Confirmation Bias”. Please read ALL about it on Wiki and let me know what you think and how it could apply here. This trap has ensnared some very smart people. Harvey1 bnjroo Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been created! NOTHING IS IMPOSSIBLE! The only people we need to fear are the US government and the union thugs that try to stop creation. Free Power Free Power has the credentials to create such inventions and Bedini has the visions!

We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
Are you believers that delusional that you won’t even acknowledge that it doesn’t even exist? How about an answer from someone without attacking me? This is NOT personal, just factual. Harvey1 kimseymd1 Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! Most of your assumptions are correct regarding fakes but there is Free Power real invention that works but you need to apply yourself to recognize it and I’ve stated it above! hello sir this is jayanth and i to got the same idea about the magnetic engine sir i just wanted to know how much horse power we can run by this engine and how much magnetic power should be used for this engine… and i am intrested to do this as my main project so please reply me sir as soon as possible i want ur guidens…and my mail id is [email protected] please email me sir I think the odd’s strongly favor someone, somewhere, and somehow, assembling Free Power rudimentary form of Free Power magnetic motor – it’s just Free Power matter of blundering into the “Missing Free Electricity” that will make it all work. Why not ?? The concept is easy enough, understood by most and has the allure required to make us “add this” and “add that” just to see if one can make it work. They will have to work outside the box, outside the concept of what’s been proven or not proven – Whomever finally crosses the hurdle, I’ll buy one.
Why not use the term over unity over perpetual motion? Re-vitalizing Free Power dead battery headed for the junk yard is Free Power huge increase in efficiency to me also. Why doesn’t every AutoZone or every auto shop have one of these? Unless the battery case is cracked every battery could be reused. The charge of Free Power re-vitalize instead of Free Power new battery. Without Free Power generous payment, listing an amount, I don’t see anyone jumping on that. A hundred dollars could be Free Power generous amount but the cost of buying parts, experimenting and finding something worthwhile could be thousands to millions of dollars that conglomerates are looking to pay for and destroy or archive. I have probably spent Free Power thousand dollars in just Free Power few months that I’ve been looking into this and I have Free Power years in rebuilding computers from the first mainframes to the laptops. I retired and now its Free Power hobby. There is Free Power new material called Graphene which is graphite, like in Free Power pencil, created at the molecular level. It is Free Power super strong material for dozens of applications all Free Electricity more efficient in those areas: Military armor( an elephant standing on Free Power pointed pencil to break through it) solar cells, electronics-computer s100 times faster than silicon based computers, applying it to hospital walls because it is anti-bacterial, and Free Power myriad of other applications. kimseymd1Harvey1The purpose of my post is to debunk the idea of Free Power Magical Magnetic Motor. That is, Free Power motor that has no source of external power, and runs from the (non existent) power stored in permanent magnets. Advances made to electric motors in the past few years are truly amazing, but are totally outside the scope of my post.

## I don’t know what to do. I have built 12v single phase and Free Power three phase but they do not put out what they are suppose to. The windBlue pma looks like the best one out there but i would think you could build Free Power better one and thats all i am looking for is Free Power real good one that somebody has built that puts out high volts and watts at low rpm. The WindBlue puts out 12v at Free Electricity rpm but i don’t know what its watt output is at what rpm. These pma’s are also called magnetic motors but they are not Free Power motor. They are Free Power generator. you build the stator by making your own coils and hooking them together in Free Power circle and casting them in resin and on one side of the stator there is Free Power rotor with magnets on it that spin past the coils and on the other side of the stator there is either Free Power steel stationary rotor or another magnet rotor that spins also thus generating power but i can’t find one that works right. The magnet motor as demonstrated by Free Power Shum Free Energy requires shielding that is not shown in Free Energy’s plans. Free Energy’s shielding is simple, apparently on the stator. The Perendev shows each magnet in the Free Energy shielded. Actually, it intercepts the flux as it wraps around the entire set of magnets. The shielding is necessary to accentuate interaction between rotor and stator magnets. Without shielding, the device does not work. Hey Gilgamesh, thanks and i hope you get to build the motor. I did forget to ask one thing on the motor. Are the small wheels made of steel or are they magnets? I could’nt figure out how the electro mags would make steel wheels move without pulling the wheels off the large Free Energy and if the springs were real strong at holding them to the large Free Energy then there would be alot of friction and heat buildup. Ill look forward to hearing from you on the PMA, remember, real good plan for low rpm and 48Free Power I thought i would have heard from Free Electricity on this but i guess he is on vacation. Hey Free Power. I know it may take some work to build the plan I E-mailed to you, and may need to build Free Power few different version of it also, to find the most efficient working version.

Why? Because I didn’t have the correct angle or distance. It did, however, start to move on its own. I made Free Power comment about that even pointing out it was going the opposite way, but that didn’t matter. This is Free Power video somebody made of Free Power completed unit. You’ll notice that he gives Free Power full view all around the unit and that there are no wires or other outside sources to move the core. Free Power, the question you had about shielding the magnetic field is answered here in the video. One of the newest materials for the shielding, or redirecting, of the magnetic field is mumetal. You can get neodymium magnets via eBay really cheaply. That way you won’t feel so bad when it doesn’t work. Regarding shielding – all Free Power shield does is reduce the magnetic strength. Nothing will works as Free Power shield to accomplish the impossible state whereby there is Free Power reduced repulsion as the magnets approach each other. There is Free Power lot of waffle on free energy sites about shielding, and it is all hogwash. Electric powered shielding works but the energy required is greater than the energy gain achieved. It is Free Power pointless exercise. Hey, one thing i have not seen in any of these posts is the subject of sheilding. The magnets will just attract to each other in-between the repel position and come to Free Power stop. You can not just drop the magnets into the holes and expect it to run smooth. Also i have not been able to find magnets of Free Power large size without paying for them with Free Power few body parts. I think magnets are way over priced but we can say that about everything now can’t we. If you can get them at Free Power good price let me know.
LoneWolffe kimseymd1 Harvey1 TiborKK Thank You LoneWolffe!! Notice how kimseymd1 spitefully posted his “Free Energy two books!.. ” spam all over this board on every one of my posts. Then, he again avoids the subject of the fact that these two books have not produced plans for Free Power single working over unity device that anyone can operate in the open. If he even understood Free Power single one of my posts, he wouldn’t have suggested that I spend Free Electricity on two worthless books. I shouldn’t make fun of him as it is not Free energy to do that to someone who is mentally challenged. I wish him well and hope that he gets the help that he obviously needs. Perhaps he’s off his meds. Harvey1: I haven’t been on here for awhile. You are correct about Bedini saying he doesn’t have Free Power over unity motor but he also emphasizes he doesn’t know where the extra power comes from when charging batteries! Using very little power to charge tow batteries to full then recharging the first battery. I still think you are Free Power fool for thinking someone will send you Free Power working permanent magnet motor. Building Free Power Bedini motor is fun and anyone can do it! I am on my third type but having problems!

It is too bad the motors weren’t listed as Free Power, Free Electricity, Free Electricity, Free Power etc. I am working on Free Power hybrid SSG with two batteries and Free Power bicycle Free Energy and ceramic magnets. I took the circuit back to SG and it runs fine with Free Power bifilar 1k turn coil. When I add the diode and second battery it doesn’t work. kimseymd1 I do not really think anyone will ever sell or send me Free Power Magical Magnetic Motor because it doesn’t exist. Therefore I’m not Free Power fool at all. Free Electricity realistic. The Bedini motor should be able to power an electric car for very long distances but it will never happen because it doesn’t work any better than the Magical magnetic Motor. All smoke and mirrors – No Working Models that anyone can operate. kimseymd1Harvey1You call this Free Power reply?

### No “boing, boing” … What I am finding is that the abrupt stopping and restarting requires more energy than the magnets can provide. They cannot overcome this. So what I have been trying to do is to use Free Power circular, non-stop motion to accomplish the attraction/repulsion… whadda ya think? If anyone wants to know how to make one, contact me. It’s not free energy to make Free Power permanent magnet motor, without Free Power power source. The magnets only have to be arranged at an imbalanced state. They will always try to seek equilibrium, but won’t be able to. The magnets don’t produce the energy , they only direct it. Think, repeating decimal…..

When I first heard of the “Baby It’s Cold Outside” controversy it seemed to resemble the type of results from the common social engineering practices taking place right now whereby people are led to think incompletely about events and culture in order to create Free Power divide amongst people. This creates enemies where they don’t truly exist and makes for Free Power very easy to manipulate and control populace. Ultimately, this leads for people to call for greater governance.
Free Energy The type of magnet (natural or man-made) is not the issue. Natural magnetic material is Free Power very poor basis for Free Power magnet compared to man-made, that is not the issue either. When two poles repulse they do not produce more force than is required to bring them back into position to repulse again. Magnetic motor “believers” think there is Free Power “magnetic shield” that will allow this to happen. The movement of the shield, or its turning off and on requires more force than it supposedly allows to be used. Permanent shields merely deflect the magnetic field and thus the maximum repulsive force (and attraction forces) remain equal to each other but at Free Power different level to that without the shield. Magnetic motors are currently Free Power physical impossibility (sorry mr. Free Electricity for fighting against you so vehemently earlier).
My older brother explained that in high school physics, they learned that magnetism is not energy at all. Never was, never will be. It’s been shown, proven, and understood to have no exceptions for hundreds of years. Something that O. U. should learn but refuses to. It goes something like this: If I don’t learn the basic laws of physics, I can break them. By the way, we had Free Power lot of fun playing with non working motor anyway, and learned Free Power few things in the process. My brother went on to get his PHD in physics and wound up specializing in magnetism. He designed many of the disk drive plates and electronics in the early (DOS) computers. bnjroo Harvey1 Thanks for the reply! I’m afraid there is an endless list of swindlers and suckers out there. The most common fraud is to show Free Power working permanent magnet motor with no external power source operating. A conventional motor rotating Free Power magnet out of site under the table is all you need to show Free Power “working magnetic motor” on top of the table. How could I know this? Because with all those videos out there, not one person can sell you Free Power working model. Also, not one of these scammers can ever let anyone not related to his scam operate the motor without the scammer hovering around. The believers are victims of something called “Confirmation Bias”. Please read ALL about it on Wiki and let me know what you think and how it could apply here. This trap has ensnared some very smart people. Harvey1 bnjroo Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been created! NOTHING IS IMPOSSIBLE! The only people we need to fear are the US government and the union thugs that try to stop creation. Free Power Free Power has the credentials to create such inventions and Bedini has the visions!
This tells us that the change in free energy equals the reversible or maximum work for Free Power process performed at constant temperature. Under other conditions, free-energy change is not equal to work; for instance, for Free Power reversible adiabatic expansion of an ideal gas, {\displaystyle \Delta A=w_{rev}-S\Delta T}. Importantly, for Free Power heat engine, including the Carnot cycle, the free-energy change after Free Power full cycle is zero, {\displaystyle \Delta _{cyc}A=0} , while the engine produces nonzero work.


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Tag Info

5

Yes, this problem is FPT. We showed this in the paper "Parameterized Edge Hamiltonicity" (Lampis, Makino, Mitsou, and Uno, DAM 2018). In particular, in this paper we state the result for Hamiltonian Path, but things doesn't change much if you want to consider Hamiltonian Cycle. In the conference version (and the slightly outdated version on the arxiv) we ...

2

distance-2 coloring is coloring in the square (d(x,y) <= 2 implies xy an edge in the square). If a graph has tw k, its square has bounded clique-width (see Gurski-Wanke and Suchan-Todinca). See Oum et al. for algorithms as in Proskurowski-Telle for graphs of bounded clique-width.

2

Yes. By Proposition 2.3 of [1], all elementary fractional extreme points of the LP correspond to subgraphs that contain odd cycles, and therefore if the graph contains no odd cycles, the LP has an optimal solution that takes on only integer values. [1] G. L. Nemhauser and L. E. Trotter Jr. Properties of vertex packing and independence system polyhedra. ...

1

Starting with 5,7,$f$ and $g$ is correct. In order to complete the maximum independent you can proceed as follows. First, delete the vertices that are reachable from an unmatched vertex via an alternating path (in your example 5,6,7,$e$,$f$,$g$). This leaves a balanced bipartite graph (that is, it has the same number of vertices on each side), and you can ...

Top 50 recent answers are included


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# Density stability; questions for those who like computer calculation

BACKGROUND: The question, which has its roots in a question asked on MO by O'Bryant, concerns the relative density of certain subsets, $B$, of ${\mathbb N}$ in congruence classes modulo a power of 2. Let $I$ be such a congruence class. I'll say that $B$ is "stable in $I$" if there is a $c$ such that $B$ has relative density $c$ in $J$ whenever $J$ is a congruence class contained in $I$ whose modulus is also a power of 2.

Suppose $B$ consists of all $n$ such that the coefficient of $x^n$ in the reciprocal of the element $g=1+x+x^4+x^9+x^{16}+\dots$ of ${\mathbb Z}/2[[x]]$ is 1. Cooper et al. showed that $B$ has density 0 in 12 of the mod 16 congruence classes. I extended the result to 3 of the 4 remaining classes. But calculations by O'Bryant suggest that in the class 15 mod 16, $B$ is stable with relative density $1/2$. For a detailed account see my note Disquisitiones Arithmeticae and online sequence A108345.

These QUESTIONS pertain to sets introduced by Cooper et. al:

1. Replace the exponents $0, 1, 4, 9, \dots$ in $g$ by the numbers $3n^2-2n$, $n \in {\mathbb Z}$, to get a new $B$. This $B$ has density 0 in 7 of the classes mod 8. Does the computer suggest that it is stable with relative density 1/2 in the class 0 mod 8?

2. Suppose the exponents are $5n^2-4n$, $n \in {\mathbb Z}$. Does the computer suggest that there's a $q$ such that the new $B$ you get is stable in each mod $q$ congruence class? And if so, what do the relative densities appear to be? (The density is provably 0 in some mod 8 classes).

3. Answer the same question as 2. when the exponents are the $5n^2-2n$, $n \in {\mathbb Z}$.

EDIT: I'll give a modified and generalized version of the question (and an expansion of my answer) using notation and ideas from my MO question on characteristic 2 thetas. Let $L$ be the field of formal Laurent series in $x$ over ${\bf Z}/2$. If $f$ (not zero) is in $L$, $B(f)$ consists of all $n$ for which the coefficient of $x^n$ in $1/f$ is 1. Now fix $l=2m+1$, $m>0$, and for $i$ in $\lbrace1,...,m\rbrace$ let $[i]$ be the element of ${\bf Z}/2[[x]]$ defined in the "thetas" question.

Question: For $q$ a power of 2, what does the computer suggest about the relative density of $B([i])$ in the various mod $q$ congruence classes? (Since all elements of $B[i]$ are congruent to $-(i^2)$ mod $l$, these relative densities are at most $1/l$).

Example: When $l=3$, it can be shown that $B([i])$ has density 0 in all congruence classes mod 8, with the possible exception of 7. And the computer (perhaps) indicates that in the 7 mod 8 class (or any class contained therein) the relative density is 1/6.

My "answer" generalizes the first sentence of the example. I made no computer calculations--indeed the computer evidence is at first sight contrary to my results because of the slow approach to zero. Let $L(q)$ contained in $L$ be the field of formal Laurent series in $x^q$. Then $L$ is the direct sum of the $(x^k)L(q)$, $k$ in $\lbrace0,...,q-1\rbrace$. Let $p_{(q,k)}$ be the obvious projection map $L\to(x^k)L(q)$. Let $S$ contained in

${\bf Z}/2[[x]]$

be the smallest ring that contains all the $[i]$ and is stable under the $p_{(q,k)}$ for all $q$ and $k$. It can be shown that every element of $S$ is the mod 2 reduction of the Fourier series of an integral weight modular form for a congruence group. A theorem of Serre then shows that if $\sum((c_n)(x^n))$ is in $S$ then the set of $n$ for which $c_n$ is 1 has density 0.

As a corollary one finds: Let $p$ be a $p_{(q,k)}$. If $p(1/[i])$ is in $S$ then $B([i])$ has density 0 in the class $k$ mod $q$.

By making use of the quintic relations from my theta question I can show that the hypothesis of the theorem holds in various cases. In particular suppose $i$ is prime to l. When $l=5$, $B=B([i])$ has density 0 in each mod 32 class except perhaps the 5 classes $n=7$ mod 8 and $n=28$ mod 32. When $l=7$, $B$ has density 0 in each mod 32 class except perhaps the 7 classes 7 mod 8, 14 mod 16 and 28 mod 32. When $l=9$, $B$ has density 0 in each mod 64 class except perhaps the 19 classes 1 and 7 mod 8, 28 mod 32, and 48 mod 64.

In the various classes qualified by "except perhaps" in the above paragraph (and the subclasses contained therein) it seems plausible that the relative densities are $1/(2l)$. But this may be wishful thinking. I hope that someone will make further calculations.

FURTHER EDIT: Heres a more explicit and more speculative version of my question. Let n_j be the negative exponents appearing in the Laurent series 1/[i], 1/[2i], 1/[4i], i/[8i],..., and q_j be the largest power of 2 dividing n_j.

QUESTION: Does computer evidence support the following speculations?

(1) The relative density of B([i]) in each congruence class n_j mod 8q_j, and in all congruence classes modulo a power of 2 contained therein, is 1/(2l).

(2) Outside of these congruence classes B([i]) has density 0.

For example when l=9 and i=1 the n_j are -16,-7,-4 and -1, and the classes in (1) are 1 mod 8, -1 mod 8, -4 mod 32 and -16 mod 128. The technique I indicated in my earlier edit shows that (2) holds in this case, so one gets 128-37 classes mod 128 where B has density 0. The technique also shows that (2) holds when l=3,5 or 7. This isn't much evidence, and there's far less for (1). But as these are the simplest answers one might hope for, I'd be interested in any calculations concerning them.

-
Hi Paul, you forgot to put brackets around the exponent $16$ in your equation g=1+x+x^4+x^9+x^16..., you should have used g=1+x+x^4+x^9+x^{16} to get the correct result $g=1+x+x^4+x^9+x^{16}...$ I don't have enough points to edit that for you, so you'll have to fix that. And you should just be able to hit the "edit" button under your question to perform the edit. MO shouldn't lose the question if you do that. –  sleepless in beantown Nov 18 '10 at 4:25
I've attempted to TeX the most recent edit. Hope I didn't corrupt the meaning. –  Gerry Myerson Nov 18 '10 at 5:23
Gerry-Thanks for the TeXing; I don't know how to do such things. The meaning is intact, though I prefer Z/2 to the ambiguous Z_2 –  paul Monsky Nov 19 '10 at 2:39
@paul, ${\bf Z}_2$ replaced by ${\bf Z}/2$. –  Gerry Myerson Nov 19 '10 at 11:31

This is really commentary on 2. and 3. of my question in the light of recent discoveries, but as I don't know how to edit the question without losing it, I'll post my discoveries as an answer. Following Cooper et. al. denote the sets B of 2. and 3. by B_(1,10) and B_(3,10). I can now prove:

a. The n in B_(1,10) that are neither 7 mod 8 nor 0 mod 32 have density 0.

b. The n in B_(3,10) that are neither 0 mod 8 nor 25 mod 32 have density 0.

I suspect that B_(1,10) is stable with relative density 1/2 in each of the 5 remaining congruence classes mod 32, and that the same is true for B_(3,10). It would in fact suffice to handle the classes 7 mod 8 for B_(1,10) and 0 mod 8 for B_(3,10), and I would appreciate calculations for these mod 8 congruence classes to confirm or disconfirm my suspicions.

The proofs of a. and b. use a deep result of Serre's on the mod J reduction of the Fourier expansion of a modular form of integral weight,when the Fourier coefficients lie in a number field, together with some simple techniques from my answers to O'Bryant's questions. I hope to post the proofs on arXiv one of these days.

EDIT: The arguments can now be found on arXiv NT 1107.4137, "The reciprocals of some characteristic 2 'theta series' ". I use the techniques mentioned above to prove the analogues to (a) and (b) when l=7,9,11,13 and 15 as well. (See the edit to my question to see what I'm describing). But I can't handle one class mod (128) for l=13 and 15, and though my method might give density 0 results in some congruence classes when l>15, the computer isn't up to the job. So I'd still be grateful to anyone willing to make further computer calculations to see whether my speculations are plausible.

-

For prime l I've now proved (a corrected version of) the speculation (2) made in the further edit to my question. (See Theorem I below). The proof avoids the extensive computer verifications made in arXiv NT 1107.4137. So let K be an algebraic closure of Z/2. Call an element g=a_0+a_1(x)+a_2(x^2)+... of K[[x]] "sparse" if the n with a_n non-zero form a set of density 0.

Lemma 1----Suppose g is in the subring R of K[[x]] generated by K and the [i]. Then g is sparse. (This follows from the fact that the elements of R are the mod 2 reductions of Fourier expansions of modular forms of integral weight, and the theorem of Serre mentioned in my previous answer).

We have shown in another question that the above ring R is the co-ordinate ring of an affine curve C. Let m_0 be the maximal ideal of R generated by [1],...,[l-1], and p_0 be the point of C corresponding to m_0. We showed in addition (using the fact that l is prime) that m_0 is the only maximal ideal of R containing any of [1],...,[l-1], and that there are (l-1)_/2 linear branches at p_0 with distinct branch tangents.

Lemma 2---Suppose g=a_0+a_1(x)+... is in K[[x]], that g is the quotient of an element of R by a product of powers of [i], and that g has positive ord at every branch of C centered at p_0. Then g is sparse. (For g is in the localization of R at every maximal ideal other than m_0. Furthermore if n is large g^n is in the localization of R at m_0. So for n large, g^n is in R, and is sparse by Lemma 1. Take n to be a large power of 2 to get the result.

---------Now let U=U_2 be the operator K[[x]]-->K[[x]] taking sum(a_n)(x^n) to sum(a_2n)(x^n). Note that U([i][i]g)=[i]U(g).

Lemma 3---The subring of Z/2[[x]] generated by the [i] is stable under U. (It suffices to show that U takes a product of terms, [ ], to an element of this ring. We argue by induction on the number of terms in the product. We may assume that the first 2 terms are [2i] and [2j]. Then [2i][2j] is the sum of [2i][j]^4, [2j][i]^4 and ([i+j][i-j])^2. Multiplying by the remaining terms in the product, applying U, and using induction we get the result).

Now let L be the field of Laurent series in x over Z/2, and L(q) be the field of Laurent series in x^q, where q is a power of 2. We write p_(q,k) for the obvious projection map L-->(x^k)L(q). Note that for g in Z/2[[x]], U(g) is the square root of p_(2,0)(g). So if g is in the ring of Lemma 3, then p_(q,0)(g) is a qth power in that ring.

Theorem I---Let q be a power of 2, and suppose that k is in {0,1,2,3,4,5,6,7}. Suppose that g_i=p_(8q,kq)(1/[i]) is in Z/2[[x]] for all i--that is to say that no negative exponents appear in any g_i. Then each g_i is sparse. (In other words each B([i]) has density 0 in each congruence class kq mod 8q).

To prove this note that p_(q,0)(1/[i]) is the quotient of p_(q,0)([i]^(8q-1)) by [i]^8q. The paragraph before Theorem I shows that this is the quotient of v^q by [i]^8q for some v in R. Applying p_(8q,kq) we find that g_i=(1/[i]^8q)(w^q), where w=p_(8,k)(v). Since p_(8,k) stabilizes R, g_i is the quotient of an element of R by a power of [i]. The exponent restriction tells us that if g is a g_j, then g has positive ord at each branch of C centered at m_0, and we invoke Lemma 2.

Example: Suppose l=13, q=16 and k=3. The negative exponents appearing in the i/[i] are -36,-23,-10, -25, -16,-9,-4,and -1. Since none of these is congruent to 48 mod 128, all the g_i are in Z/2[[x]]. It follows from Theorem I that each B([i]) has density 0 in the congruence class 48 mod 128, a result that had eluded me.

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# statement logic, negation

• January 29th 2009, 01:53 AM
chtslip
statement logic, negation
Hi there, I need help with this one
Please provide the solution and explain
(I can only do simple statements myself, so I'm stumped here)

Negate the following statement:
$\forall$x $\in$R $\exists$y $\in$R, x+y=0
(for all real valued x, there's some real valued y, where x+y=0)

Any help is appreciated
Thank you
• January 29th 2009, 03:16 AM
Plato
Here is the general rule: $\neg \left( {\forall x} \right)\left( {\exists y} \right)P(x,y) \equiv \left( {\exists x} \right)\left( {\forall y} \right)\neg P(x,y)$


text:
• 研究报告 •

农村居民点时空变化特征及驱动力分析——以宁波市北仑区为例

1. (1宁波大学地理与空间信息技术系， 浙江宁波 315211；2浙江省海洋文化与经济研究中心， 浙江宁波 315211)
• 出版日期:2018-02-10 发布日期:2018-02-10

Spatialtemporal changes and driving forces of rural settlements: A case study of Beilun District, Ningbo.

FENG Bai-xiang1, LI Jia-lin1,2*, HE Gai-li1, HUANG Ri-peng1

1. (1Department of Geography & Spatial Information Techniques, Ningbo University, Ningbo 315211, Zhejiang, China; 2Research Center for Marine Culture and Economy, ZhejiangProvince, Ningbo 315211, China).
• Online:2018-02-10 Published:2018-02-10

Abstract: Beilun, a district of Ningbo City, is located in the southeastern coastal area of China. Based on six periods TM/OLI remote sensing images of 1990, 1995, 2000, 2005, 2010 and 2015, we applied landuse dynamic index, conversion matrix, kernel density estimation, average nearest neighbor index and landscape pattern index to analyze the spatialtemporal dynamics and driving forces of rural settlements in the last 25 years in Beilun District. The results showed that the landuse dynamic index was generally positive in rural settlements, and the residential land showed a slow expansion trend. The average patch area of rural residential area in Beilun increased in the last 25 years, with an average annual growth rate of 2.51%. In the first twoperiods studied here, the average patch size was mainly affected by their number. In the later two periods, the total number of patch and the total patch area were the drivers. At the last period, the total area of the patches was the dominant driving factor. The differences of patch size among rural settlements were more and more significant. Agricultural land and forest land were the dominant types of land conversion in the settlements, especially for agricultural land. The urban construction had no obvious impacts on the settlements. The spatial patterns of rural settlements were in agglomeration in the west, dualcore interaction in center and scattered distribution in the east. In 2015, there was an obvious highdensity area near the northern islands (Daxie). Theoverall distribution of the settlements was changed from dispersed pattern into clusters, with the degree of agglomeration continuing to increase. The distribution of rural settlements in Beilun was affected by population density, road, landscape and policy. Population growth significantlyaffected the spatial patterns of rural settlements in 1995-2000 and 2000-2005. The directions of state roads and county roads were closely linked with the spatial distribution of settlements.Policies can promote the changes of distribution of settlements and lay the foundation for the future distribution. The settlements were mostly distributed in the low-lying area  (DEM <100 m).


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# Trigonometric Functions Of Any Angle

All new and solved questions in Trigonometric Functions Of Any Angle category

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##### How do you find the exact value of  tan (pi/3) +cos pi?
How do you find the exact value of  tan (pi/3) +cos pi?...


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text:
# Tag Info

77

No questions are required!

45

(a) I claim that the expected typing length are the same for both monkeys. I guess something in my argument will be incorrect, as jafe's answer has 9 approvals, but finding that incorrectness would be helpful to me. I prove that the probabilty that a monkey ends his typing after exactly $n$ letters is the same for both monkeys, which then implies that the ...

30

(a) Edit: This is incorrect, see comments

28

OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that. The simplest somewhat-plausible way to handle #1 ...

25

Monkey problem To settle down which monkey is faster on average, I'll use Markov chains and Mathematica. Define a state $i$, for $i = 0..6$, as that the monkey 1 has currently written $i$ correct subsequent characters of COCONUT, but has never written all of COCONUT. Define a state 7 as the monkey 1 having written COCONUT at least once. Since state 7 cannot ...

23

The answer is Proof Alternative proof

23

21

A difference between the Monty Hall scenario and the train platform scenario is that After you remember that the train will definitely not be leaving from platform 2,

16

I'd Then Then So

16

(a) has been adequately answered by In (b), though: Consider: In this case: Now consider:

16

This gambling problem is the famous St. Petersburg paradox. It is a paradox because The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the ...

15

@Ingix provides a good proof, one certainly worthy of the tick, but one that could perhaps use some reinforcement. The answer is that the N number of letters would have the same expected value. A few points that might help @jafe realise the flaw in his proof. The order of independent events occurs has no directional bias. For example, drawing a King Of ...

15

13

I have a solution with a success rate of 93.5%, according to my simulations. The reason this solution works so well is Here's my code that I used to verify my solution:

12

For clarify: No matter who you ask, he will answer the correct door.

11

There is a chance that the other jar is empty. This puzzle is asking the equivalent of which can be calculated by

11

Suppose $t(n)$ is the average number of spins you get if you start with $\$10n$. We want$t(2)$. If you start with$n$ten-dollar bills, put$n-1$in your pocket and play until you are broke, then take the next \$10 out and play until you are broke, and so on: this is exactly equivalent to just starting with $\$10n$and playing until you're broke, so$t(n)=...

11

I would pay

10

Originally there are; 10 S, 6 V and 5C For the first case where he wants to have 7 S, 5 V and , 4 C. (so we dont want to have 3S, 1V and 1C) In the worst case scenario; so I do not want to continue for the rest since the same methodology works for them too: This will give you the number of cookies to guarantee to have some specific number of cookies....

10

Here's how the original Monty Hall worked: You pick a door. (2/3 chance of it being a goat, 1/3 chance of it being a car) Monty Hall opens a door that he knows has a goat in it, and then asks you if you'd like to switch to the other door. 2/3 of the time, the door Monty Hall opened is the only other door with a goat - because 2/3 of the time, you picked a ...

10

To do this, Finally, I can give you the Python code if you want. Not sure it can be spoilered (I've never learned how to have multi-line spoilers). Python Code

9

Proceeding along the same lines as hexomino, I get a different result. The expectation The place where I (ha!) diverge from hexomino's answer is We can try to deal with this inconvenient situation in three ways. First, Second, Third, The upshot of all this is Credit where due: hexomino did a lot of this before I did (with, unfortunately, an error, but ...

9

Ask the following question of all three guards: Now the number of Yeses (Y) will be between 0 and 3 inclusive. If Y=1, go through that door. The position may either be 1. in which case you go to heaven, or it may be one of in which case you go to hell. If Y=2, namely 3. then pick one of the Yeses at random and ask the utterer the same question again. ...

9

You should because of In numbers:

8

First of all, let's do the obvious calculation. Now note that Hence But of course So the answer to the question as stated is But it's important to notice One common way to deal with the second of these is Let's consider both of those together In fact

8

One strategy: Probability of success:

8

There are several nuances to this question. First of all, it asks how much you are willing to pay, not what price is fair. Second, you have to understand, that even if a game is fair, that does not mean that It is reasonable to play it. For example, if someone offers me a one in a million chance to win a million dollars for 1, I will take it. It seems ...

8

I would not pay anything. I would not play. I would encourage you to not play. Are you doing okay? I'm willing to help you out of you need help. I would offer you a hug. You are my best friend, and you live right next to me. Any outcome of this game that would be monetarily meaningful to either of us would also most likely be highly damaging to our ...

7

There is a clever trick to finding the kind of expected times that appear in part (a). It turns out that both for COCONUT and TUNOCOC, the expected time is just $26^7$ (so the monkeys have equal expected times to win), but for this to be convincing, I will demonstrate the method on a case where the expected value is unexpected: the word ABRACADABRA. Suppose ...

7

N=1,677 Even with all of the complex mathmatics being done, probability only gives us a way to calculate the answer devoid of other information. Since you gave an example, but didn't specifically say it WASN'T your example, the most likely hypothesis is that the site you listed as an example was the site you visited. Random.cat actually lists directly on ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.

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What if you were given a circle with a quadrilateral inscribed in it? How could you use information about the arcs formed by the quadrilateral and/or the quadrilateral's angle measures to find the measure of the unknown quadrilateral angles? After completing this Concept, you'll be able to apply the Inscribed Quadrilateral Theorem to solve problems like this one.

### Guidance

An inscribed polygon is a polygon where every vertex is on a circle. Note, that not every quadrilateral or polygon can be inscribed in a circle. Inscribed quadrilaterals are also called cyclic quadrilaterals. For these types of quadrilaterals, they must have one special property. We will investigate it here.

Tools Needed: pencil, paper, compass, ruler, colored pencils, scissors

1. Draw a circle. Mark the center point $A$ .
2. Place four points on the circle. Connect them to form a quadrilateral. Color the 4 angles of the quadrilateral 4 different colors.
3. Cut out the quadrilateral. Then cut the quadrilateral into two triangles, by cutting on a diagonal.
4. Line up $\angle B$ and $\angle D$ so that they are adjacent angles. What do you notice? What does this show?

This investigation shows that the opposite angles in an inscribed quadrilateral are supplementary. By cutting the quadrilateral in half, through the diagonal, we were able to show that the other two angles (that we did not cut through) formed a linear pair when matched up.

Inscribed Quadrilateral Theorem: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary.

#### Example A

Find the value of the missing variable.

$x+80^\circ=180^\circ$ by the Inscribed Quadrilateral Theorem. $x=100^\circ$ .

$y+71^\circ=180^\circ$ by the Inscribed Quadrilateral Theorem. $y=109^\circ$ .

#### Example B

Find the value of the missing variable.

It is easiest to figure out $z$ first. It is supplementary with $93^\circ$ , so $z=87^\circ$ . Second, we can find $x$ . $x$ is an inscribed angle that intercepts the arc $58^\circ + 106^\circ = 164^\circ$ . Therefore, by the Inscribed Angle Theorem, $x=82^\circ$ . $y$ is supplementary with $x$ , so $y=98^\circ$ .Find the value of the missing variables.

#### Example C

Find $x$ and $y$ in the picture below.

The opposite angles are supplementary. Set up an equation for $x$ and $y$ .

$(7x+1)^\circ+105^\circ &= 180^\circ && (4y+14)^\circ+(7y+1)^\circ = 180^\circ\\7x+106^\circ &= 180^\circ && \qquad \qquad \quad \ 11y+15^\circ = 180^\circ\\7x &= 84^\circ && \qquad \qquad \qquad \quad \quad 11y = 165^\circ\\x &= 12^\circ && \qquad \qquad \qquad \qquad \quad y = 15^\circ$

Watch this video for help with the Examples above.

### Vocabulary

A circle is the set of all points that are the same distance away from a specific point, called the center . A radius is the distance from the center to the circle. A chord is a line segment whose endpoints are on a circle. A diameter is a chord that passes through the center of the circle. The length of a diameter is two times the length of a radius. A central angle is an angle formed by two radii and whose vertex is at the center of the circle. An inscribed angle is an angle with its vertex on the circle and whose sides are chords. The intercepted arc is the arc that is inside the inscribed angle and whose endpoints are on the angle. An inscribed polygon is a polygon where every vertex is on the circle.

### Guided Practice

Quadrilateral $ABCD$ is inscribed in $\bigodot E$ . Find:

1. $m\angle A$
2. $m\angle B$
3. $m\angle C$
4. $m\angle D$

First, note that $m\widehat{AD}=105^\circ$ because the complete circle must add up to $360^\circ$ .

1. $m\angle A=\frac{1}{2}m\widehat{BD}=\frac{1}{2}(115+86)=100.5^\circ$

2. $m\angle B=\frac{1}{2}m\widehat{AC}=\frac{1}{2}(86+105)=95.5^\circ$

3. $m\angle C=180^\circ-m\angle A=180^\circ-100.5^\circ=79.5^\circ$

4. $m\angle D=180^\circ-m\angle B=180^\circ-95.5^\circ=84.5^\circ$

### Practice

Fill in the blanks.

1. A $(n)$ _______________ polygon has all its vertices on a circle.
2. The _____________ angles of an inscribed quadrilateral are ________________.

Quadrilateral $ABCD$ is inscribed in $\bigodot E$ . Find:

1. $m\angle DBC$
2. $m \widehat{BC}$
3. $m \widehat{AB}$
4. $m\angle ACD$
5. $m\angle ADC$
6. $m\angle ACB$

Find the value of $x$ and/or $y$ in $\bigodot A$ .

Solve for $x$ .

Use the diagram below to find the measures of the indicated angles and arcs in problems 14-19.

1. $m \angle EBO$
2. $m \angle EOB$
3. $m \widehat{BC}$
4. $m \angle ABO$
5. $m \angle A$
6. $m \angle EDC$

### Vocabulary Language: English Spanish

central angle

central angle

An angle formed by two radii and whose vertex is at the center of the circle.
chord

chord

A line segment whose endpoints are on a circle.
diameter

diameter

A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
inscribed angle

inscribed angle

An angle with its vertex on the circle and whose sides are chords.
intercepted arc

intercepted arc

The arc that is inside an inscribed angle and whose endpoints are on the angle.
Inscribed Polygon

Inscribed Polygon

An inscribed polygon is a polygon with every vertex on a given circle.

The Inscribed Quadrilateral Theorem states that a quadrilateral can be inscribed in a circle if and only if the opposite angles of the quadrilateral are supplementary.

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle.


text:
## Nagoya Mathematical Journal

### Existence of extremal Beltrami coefficients with nonconstant modulus

Guowu Yao

#### Abstract

Suppose that $[\mu]_{T(\Delta)}$ is a point of the universal Teichmüller space $T(\Delta)$. In 1998, Božin, Lakic, Marković, and Mateljević showed that there exists $\mu$ such that $\mu$ is uniquely extremal in $[\mu]_{T(\Delta)}$ and has a nonconstant modulus. It is a natural problem whether there is always an extremal Beltrami coefficient of constant modulus in $[\mu]_{T(\Delta)}$ if $[\mu]_{T(\Delta)}$ admits infinitely many extremal Beltrami coefficients; the purpose of this paper is to show that the answer is negative. An infinitesimal version is also obtained. Extremal sets of extremal Beltrami coefficients are considered, and an open problem is proposed. The key tool of our argument is Reich’s construction theorem.

#### Article information

Source
Nagoya Math. J., Volume 199 (2010), 1-14.

Dates
First available in Project Euclid: 14 September 2010

Permanent link to this document
https://projecteuclid.org/euclid.nmj/1284471568

Digital Object Identifier
doi:10.1215/00277630-2010-001

Mathematical Reviews number (MathSciNet)
MR2730409

Zentralblatt MATH identifier
1227.30021

#### Citation

Yao, Guowu. Existence of extremal Beltrami coefficients with nonconstant modulus. Nagoya Math. J. 199 (2010), 1--14. doi:10.1215/00277630-2010-001. https://projecteuclid.org/euclid.nmj/1284471568

#### References

• [1] V. Božin, N. Lakic, V. Marković, and M. Mateljević, Unique extremality, J. Anal. Math. 75 (1998), 299–338.
• [2] C. J. Earle and N. Lakic, Variability sets on Riemann surfaces and forgetful maps between Teichmüller spaces, Ann. Acad. Sci. Fenn. Math. 27 (2002), 307–324.
• [3] J. Fan and J. Chen, On the equivalence of extremal Teichmüller mapping, Sci. China Ser. A. 52 (2009), 77–86.
• [4] V. Mateljević and V. Marković, The unique extremal QC mapping and uniqueness of Hanh-Banach extensions, Mat. Vesnik 48 (1996), 107–112.
• [5] E. Reich, A criterion for unique extremality of Teichmüller mappings, Indiana Univ. Math. J. 30 (1981), 441–447.
• [6] E. Reich, The unique extremality counterexample, J. Anal. Math. 75 (1998), 339–347.
• [7] E. Reich, Non-uniquely extremal quasiconformal mappings, Libertas Math. 20 (2000), 33–38.
• [8] E. Reich and K. Strebel, “Extremal quasiconformal mappings with given boundary values” in Contributions to Analysis: A Collection of Papers Dedicated to Lipman Bers, Academic Press, New York, 1974, 375–391.
• [9] K. Strebel, Point shift differentials and extremal quasiconformal mappings, Ann. Acad. Sci. Fenn. Math. 23 (1998), 475–494.
• [10] G. W. Yao, Is there always an extremal Teichmüller mapping? J. Anal. Math. 94 (2004), 363–375.
• [11] G. W. Yao and Y. Qi, On the modulus of extremal Beltrami coefficients, J. Math. Kyoto Univ. 46 (2006), 235–247.
• [12] Z. Zhou, J. Chen, and Z. Yang, On the extremal sets of extremal quasiconformal mappings, Sci. China Ser. A 46 (2003), 552–561.


text:
# Quantum register

In quantum computing, a quantum register is a system comprising multiple qubits[1]. It is the quantum analog of the classical processor register. Quantum computers perform calculations by manipulating qubits within a quantum register.

## Definition

An ${\displaystyle n}$ size quantum register is a quantum system comprising ${\displaystyle n}$ qubits.

The Hilbert space, ${\displaystyle {\mathcal {H}}}$, in which the data stored in a quantum register is given by ${\displaystyle {\mathcal {H}}={\mathcal {H_{n-1}}}\otimes {\mathcal {H_{n-2}}}\otimes \ldots \otimes {\mathcal {H_{0}}}}$.[2]

## Quantum vs. Classical Register

First, there's a conceptual difference between the quantum and classical register. An ${\displaystyle n}$ size classical register refers to an array of ${\displaystyle n}$ flip flops. An ${\displaystyle n}$ size quantum register is merely a collection of ${\displaystyle n}$ qubits.

Moreover, while an ${\displaystyle n}$ size classical register is able to store a single value of the ${\displaystyle 2^{n}}$ possibilities spanned by ${\displaystyle n}$ classical pure bits, a quantum register is able to store all ${\displaystyle 2^{n}}$ possibilities spanned by quantum pure qubits in the same time.

For example, consider a 2-bit-wide register. A classical register is able to store only one of the possible values represented by 2 bits - ${\displaystyle 00,01,10,11\quad (0,1,2,3)}$ accordingly.

If we consider 2 pure qubits in superpositions ${\displaystyle |a_{0}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )}$ and ${\displaystyle |a_{1}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle -|1\rangle )}$, using the quantum register definition ${\displaystyle |a\rangle =|a_{0}\rangle \otimes |a_{1}\rangle ={\frac {1}{2}}(|00\rangle -|01\rangle +|10\rangle -|11\rangle )}$ it follows that it is capable of storing all the possible values spanned by two qubits simultaneously.

## References

1. ^ Ekert, Artur; Hayden, Patrick; Inamori, Hitoshi (2008). "Basic concepts in quantum computation". arXiv:quant-ph/0011013.
2. ^ Major, Günther W., V.N. Gheorghe, F.G. (2009). Charged particle traps II : applications. Berlin: Springer. p. 220. ISBN 978-3540922605.


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# Speed of enumerating integer points in polytope

I'm trying to enumerate all integer points in some polytopes, but sage takes way too long.

This is a pretty minimal example of one of these polytopes.

q = MixedIntegerLinearProgram(maximization=False)
x = q.new_variable(integer=True, nonnegative=True)

q.add_constraint(88 * x[2] == -306 * x[0] + 1 * x[1])
q.add_constraint( 1 * x[3] == 42636 * x[0] + 42 * x[1])
q.add_constraint(11 * x[4] == 30804 * x[0] + -42 * x[1])

P = q.polyhedron()

P
# A 1-dimensional polyhedron in QQ^5 defined as the convex hull of 2 vertices

P.Vrepresentation()
# (A vertex at (1, 34/7, 5134/7, 73440, 0), A vertex at (1, 0, 306, 55488, 1632))

Now, I would like to enumerate the integral points in this (1-dimensional) polytope. Looking at the second coordinate of the two boundary points, it is clear there can be at most five such points (in fact, there are exactly five).

However, P.integral_points() seems to take forever.

Is there anything I can do to speed this up? I would like to do similar problems that are higher-dimensional, and have a lot more constraints (bounding hyperplanes).

Earlier, I used a MILP library (COIN) to solve the system, giving one integer point, then further constrain the system to exclude this solution, solve again, etc. This is fast enough for this particular example, but becomes unfeasible when the dimension of the polytope and/or the number of integer points increases.

edit retag close merge delete

Maybe you can find one point, shift it to zero and then work on the subspace spanned by the polytope ? That would make the ambient dimension smaller.

( 2014-10-15 13:48:42 -0500 )edit

Sort by » oldest newest most voted

In Sage 8.1, I can reproduce your code as follows (notice that I added "ppl" as a solver since the default behavior changed between your post and this answer).

sage: q = MixedIntegerLinearProgram(maximization=False, solver="ppl")
....: x = q.new_variable(integer=True, nonnegative=True)
....:
....: q.add_constraint(88 * x[2] == -306 * x[0] + 1 * x[1])
....: q.add_constraint( 1 * x[3] == 42636 * x[0] + 42 * x[1])
....: q.add_constraint(11 * x[4] == 30804 * x[0] + -42 * x[1])

In order to speed things up, I use normaliz.

(You can install it by typing the commands in a terminal:

sage -i normaliz
sage -i pynormaliz

)

and then use the backend normaliz for the polyhedron:

....: P = q.polyhedron(backend="normaliz")
....: P.integral_points()
....:
((1, 0, 306, 55488, 1632),
(1, 1, 394, 59184, 1296),
(1, 2, 482, 62880, 960),
(1, 3, 570, 66576, 624),
(1, 4, 658, 70272, 288))

which should appear instantaneously.

more

try using Latte Integrale (an experimental package in Sage), called latte_int; it implements Barvinok's algorithm for integer point counting. There is however no interface between Latte and Sage developed, so you need to create files to be real into Latte...

more

Unfortunately, it seems Latte Integrale does not enumerate the lattice points, it only counts them. I need the coordinates of all the integer points, for further calculations.

( 2014-10-13 06:27:45 -0500 )edit

it does count them, in a way; you have to call count with the option --multivariate-generating-function; such a function "simplifies" to a (Laurent) polynomial, with exponent vectors of the monomials corresponding to the integer points.

( 2014-10-13 10:07:10 -0500 )edit

I see. It does not sound like a very convenient way to represent the thousands of points in somewhat higher-dimensional spaces of my other cases. I'll give it a try, though. Thanks for the suggestion!

( 2014-10-13 11:29:33 -0500 )edit

it is the most compact way known, and this is what counts.IIRC, the rational function consists of summands, each of them a rational function corresponding to a simplex, simplices forming a kind of partition of the polytope.So one can probably only take summands corresponding to x^K with positive signs, if you don't mind getting the same point several times.

( 2014-10-14 02:16:57 -0500 )edit

I'm trying to do an intermediately big case, where the dimension is 8, and the polytope has 1804 vertices. Latte Integrale has already been computing for over 20 minutes (just counting the points, for now), and there is no indication how long it will take. It just says: "431000 unimodular cones done", this number increasing by a 1000 every few seconds, but I don't know how many unimodular cones it must do.

( 2014-10-14 06:40:05 -0500 )edit


text:
# Singularity analysis: logarithmic scale with general alpha and integral beta (e.g. harmonic numbers)

I'm trying to do singularity analysis to get coefficient asymptotics on functions of the following type:

$$(1-z)^{-\alpha} \log^k \frac{1}{1-z}$$

where $\alpha \notin \mathbb{Z}_{\leq 0}$ general and

$k \in \mathbb{Z}_{\geq 0}$ integral.

The book "Analytic combinatorics" from Flajolet and Sedgewick says that for this special case there exists an complete asymptotic expansion of the form

$$[z^n] (1-z)^{-\alpha} \log^k \frac{1}{1-z} \sim \frac{n^{\alpha-1}}{\Gamma(\alpha)} \left( E_0(\log n) + \frac{E_1(\log n)}{n} + \dots \right)$$

where $E_j$ are polynomials of degree $k$ (see http://algo.inria.fr/flajolet/Publications/AnaCombi/book.pdf page 387, page 403 in pdf).

How can i calculate this $E_j$ explicitly to get the complete asymptotic expansion?

-


text:
# How do you find the sum of (c-4)+(3c+9)?

Dec 1, 2016

$4 c + 5$

#### Explanation:

Step 1) Remove the parenthesis:

$c - 4 + 3 c + 9$

Step 2) Rearrange to combine like terms:

$c + 3 c - 4 + 9$

$\left(1 c + 3 c\right) + \left(- 4 + 9\right)$
$\left(1 + 3\right) c + 5$
$4 c + 5$


text:
# Notation For Dynamics#

What is the Newtonian Notation for velocity and acceleration?

## Part 1#

Choose the best answer.

• Velocity is $$v$$, $$\dot x$$; Acceleration is $$a$$, $$\dot v$$, $$\ddot x$$.

• Velocity is $$\dot v$$; Acceleration is $$\dot a$$.

• Velocity is $$\ddot v$$, $$\dot x$$; Acceleration is $$\ddot a$$, $$\dot a$$.

• Velocity is $$\ddot x$$, $$v$$; Acceleration is $$\dot x$$, $$\ddot v$$, $$a$$.


text:
Warning

This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.

# node_connected_component¶

node_connected_component(G, n)[source]

Return the nodes in the component of graph containing node n.

Parameters: G (NetworkX Graph) – An undirected graph. n (node label) – A node in G comp – A set of nodes in the component of G containing node n. set

Notes

For undirected graphs only.


text:
# AoC Day 15 – Rambunctious Recitation

This is a solution to Advent of Code 2020 day 15, written in Raku.

### Part One

Given your starting numbers, what will be the 2020th number spoken?

#### Raku

This a quick-n-dirty solution in Raku. I nearly messed up pushing the last starting input to the %seen list before considering it.

  my @input = 2, 1, 10, 11, 0, 6;

my %seen;
my $last = @input.pop; for ^@input.elems ->$i {
%seen{@input[$i]} =$i;
}

for @input.elems..^2019 -> $i { my$new;
if %seen{$last}:exists {$new = $i - %seen{$last};
} else {
$new = 0; } %seen{$last} = $i;$last = $new; } say$last;
232

### Part Two

Given your starting numbers, what will be the 30000000th number spoken?

#### Raku

The solution to part one was efficient enough that it worked for part two as well. A bit slow to run but manageable.

  my @input = 2, 1, 10, 11, 0, 6;

my %seen;
my $last = @input.pop; for ^@input.elems ->$i {
%seen{@input[$i]} =$i;
}

for @input.elems..^(30000000-1) -> $i { my$new;
if %seen{$last}:exists {$new = $i - %seen{$last};
} else {
$new = 0; } %seen{$last} = $i;$last = $new; } say$last;
18929178


text:
Aggregation of Marginal Gains

If you want to get better at something, you need a plan. Improvement doesn’t happen on its own. But once you have that plan, a bigger challenge is executing on it along with your other responsibilities.

One way to increase your chances of following through on changes is not to try to make big changes all at once. Instead, make small changes, but make them regularly. Let’s see how that works.

« Continue »

Programmer Skills (and Salaries) According to Stack Exchange

In July of this year, Stack Exchange Inc. released an online tool that lets you calculate how much money you would make if you worked there. The number you get out of the tool is based on four factors. There’s a salary floor based on the position you select (e.g., Developer or Product Designer), an adjustment based on your years of professional experience, and a bonus for living in one of a few high cost cities (New York, San Francisco, or London). Finally, the tool takes into account your skills.

Having written in the past about skills for programmers, I was interested to see what Stack Exchange decided was important for success in a programming job. Here’s what I found.

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Getting Past a Competitive Programming Plateau

In the Peak book, the authors describe the following learning challenge in a section called “Getting Past Plateaus”:

When you first start learning something new, it is normal to see rapid — or at least steady — improvement, and when that improvement stops, it is natural to believe you’ve hit some sort of implacable [immovable] limit. So you stop trying to move forward, and you settle down to life on that plateau. This is the major reason that people in every area stop improving.

The concept of the learning plateau is one way to describe how people approach learning, work, and self-improvement. With a new skill, there’s an initial period of excitement driven by how easy it is to make progress. Then a plateau arrives, and you have to decide whether to push through it or stick with your current skill level. And even if you push through it, you can look forward to another plateau where you’ll get to make the same decision again.

Skill Components

For most skills in life, there’s no need to push through plateau after plateau. It’s not worth the effort to become an expert in driving a car unless that’s your hobby or profession. But for the field that you specialize in, you probably do want to keep getting better. So what’s the secret to avoiding or conquering a learning plateau? The authors of Peak have something to say about that as well:

Any reasonably complex skill will involve a variety of components, some of which you will be better at than others. Thus, when you reach a point at which you are having difficulty getting better, it will be just one or two of the components of that skill, not all of them, that are holding you back.

According to this approach, the way to resist the plateau effect is to break down your target skill into its constituent parts, and be prepared to target those parts individually. In the book, the authors use typing speed as an example. Everyone who learns to type eventually reaches a speed plateau. Physical constraints mean you can’t keep increasing your typing speed forever. But you may plateau at a speed that is below your physical limits, or at least is slower than you want.

One idea for increasing your typing speed is just to push yourself to type faster whenever you get the chance. But according to the authors, there’s a more effective way. Rather than trying to type faster 100% of the time, try typing faster for just 15-20 minutes per day. During that time, document the mistakes you make. It’s likely that some letters or letter combinations will trip you up more than others. Once you identify them, you can more efficiently target those components, rather than trying to get better at the skill all at once.

Typing happens to be one component of the skill known as competitive programming. If your typing skills are slower than average, or if you’re competing at a high level in timed contests, working on your typing speed might be worthwhile. But for most competitive programming enthusiasts, working on other skills is more likely to produce results. What are those other skills?

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Three Ways to Solve UVa 108

UVa 108 is rated as a Level 1 (easy) problem by uHunt, but its solution nevertheless contains some interesting techniques. Here’s a summary of the problem statement:

Given an $N \times N$ array $A$ of positive and negative integers, print the sum of the nonempty subarray of $A$ that has the maximum sum. The sum of a subarray is defined as the sum of its elements.

uHunt lists this problem in the section called Max 2D Range Sum, a subcategory of Dynamic Programming. But before we get into the dynamic programming solution, let’s examine the Complete Search approach.

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text:
Which of the following choices accurately describe what makes one person different from another with regard...

Question:

Which of the following choices accurately describe what makes one person different from another with regard to traits (such as height) that have a very high heritability?

Choose all that apply (can be more than 1 option, or maybe its just 1)

Most of us have exactly the same genes and alleles, and it’s the environment we were raised in that makes us different

Random factors during development is the most significant causes of differences between people

What makes us different is not that we have different genes, but that we have different alleles of many of those genes

None of the other choices are accurate

What makes us different from each other is that we have different genes from each other


text:
Tickets numbered from 1 to 12 are mixed up together, and then a ticket is withdrawn at random.
Question:

Tickets numbered from 1 to 12 are mixed up together, and then a ticket is withdrawn at random. Find the probability that the ticket has a number which is a multiple of 2 or 3.

Solution:

We know that,

Probability of occurrence of an event

$=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$

Total no. of outcomes are 12

Desired output is picking a number which is multiple of 2 or 3 . So, desire outputs are 2 , $3,4,6,8,9,10,12$. Total no.of desired outputs are 8

Therefore, the probability of getting a number which is multiple of 2 or 3

$=\frac{8}{12}$

$=\frac{2}{3}$

Conclusion: Probability of picking a ticket which is multiple of 2 or 3 is $\frac{2}{3}$


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# help

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Let $$\mathbf{x}$$ and $$\mathbf{y}$$ be vectors such that $$\operatorname{proj}_{\mathbf{x}}(\mathbf{y})=\dbinom{5}{3}$$ and $$\operatorname{proj}_{\mathbf{y}}(\mathbf{x})=\dbinom{2}{-2}$$. Compute the ratio $$\dfrac{\|\mathbf{x}\|}{\|\mathbf{y}\|}$$.

Feb 28, 2020

#1
+24366
+1

Let $$x$$ and $$y$$ be vectors such that $$\operatorname{proj}_{\mathbf{x}}(\mathbf{y})=\dbinom{5}{3}$$and $$\operatorname{proj}_{\mathbf{x}}(\mathbf{y})=\dbinom{5}{3}$$.
Compute the ratio $$\dfrac{\|\mathbf{x}\|}{\|\mathbf{y}\|}$$.

$$\begin{array}{|rcll|} \hline \cos(\varphi) = \dfrac{\sqrt{8}}{\|\mathbf{x}\|} &=& \dfrac{\sqrt{34}}{\|\mathbf{y}\|} \\\\ \dfrac{\sqrt{8}}{\|\mathbf{x}\|} &=& \dfrac{\sqrt{34}}{\|\mathbf{y}\|} \\\\ \dfrac{\|\mathbf{x}\|}{\|\mathbf{y}\|} &=& \dfrac{\sqrt{8}}{\sqrt{34}} \\ \hline \end{array}$$

Feb 28, 2020


text:
1,104 views

The minimum number of nodes (both leaf and non-leaf) of $B^{+}$ tree index required for storing $5500$ keys and order of $B^{+}$ tree is $8$________________(order is max pointers a node can have)

See here first level should be divide by $7$

$2nd$ levelshould divide by only $8$ or $7\times 8$

because, each $7$ pointer of 1st level has $8$ pointer in 2nd level.

Am I missing something??

But in ans they divided by only $8$ :(

### 4 Comments

but why dividing by $8$ and not $7*8$ in 2nd level??

2nd level has $56$ pointers

rt??
Getting total levels as 5 and total nodes as 903. Is that right?
edited by

Do it for small value and draw it @srestha

lets say we have 10 keys and order is 3.

so keys in each node is 3-1=2.

now in leaf node we divide by 3 or 2 ? ....we divide by 2

level 1

10/2 = 5 nodes. (5 nodes each having 2 keys and 1 pointer pointing to its right node.)

level 2

now each of the 5 nodes can be pointed of by 1 pointer

1st node -> 2 value and 3 pointer

2nd node -> 1 value and 2 pointer.

so we need 5 pointers and each node can have  3 pointers

So we need 2 nodes

level 3

now each of these 2 nodes can be pointed of by 1 pointer

1st node -> 1 key 2 pointer. = root node

So leaf level is divided by (keys in each node) and non leaf by (child pointers of each node.)

## 2 Answers

Best answer

Sorry,Total nodes = 901

Ceil(5500/7) = ceil(785.71) = 786 --> Level 1

ceil(786/8) = ceil(98.25) = 99 --> Level 2

ceil(99/8) = ceil(12.375) = 13 -->Level 3

ceil(13/8) = ceil(1.625) = 2 --> Level 4

ceil(2/8) = ceil(0.25) = 1 -->  Level 5

Minimum number of nodes = 786 + 99 +13 +2 +1

= 901

Please refer this : https://gateoverflow.in/163103/b-tree-with-sparse-dense-indexing

It will help to understand why 8 and not 7.

by

As per Navathe “pleaf to denote the order for leaf nodes, which we define as being the maximum number of data pointers in a leaf node.”

Also see this for reference https://gateoverflow.in/269367/b-tree-self-doubt

As per question only one order is mentioned hence we take Pleaf also as 8 and therefore 8 keys are present in each leaf node.

Ceil(5500/8) = ceil(687.5) = 688 --> Level 4 (5500 divided by keys per node)

ceil(688/8) = ceil(86) = 86 --> Level 3 (here we are dividing by 8 and not 7 because block pointer are 8 which takes us to next level)

ceil(86/8) = ceil(10.75) = 11 -->Level 2

ceil(11/8) = ceil(1.375) = 2 --> Level 1

ceil(2/8) = ceil(0.25) = 1 -->  Level 0

Minimum number of nodes =  688 + 86 +11 +2 +1 = 788

Total nodes = 788

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text:
By Year By Month By Week Today Search Jump to month January February March April May June July August September October November December 2017 2020 2021 2022 2023 2024 2025 2026 2027
Seminario de Álgebra

Seminario de Álgebra

Viernes 7 de julio a las 11:30 en el aula 520 del módulo 17

Conferenciante: Ya'acov Peterzil (Univesity of Haifa)

Título: The topological closure of algebraic and o-minimal flows in compact tori

Abstract: Let p:C^n ->A be the covering map of a complex abelian variety and let X be an algebraic sub-variety of C^n, or more generally a definable set in an o-minimal expansion of the real field. Ullmo and Yafaev investigated the topological closure of p(X) in A in the above two settings and conjectured that the frontier of p(X) can be described, when X is algebraic, as finitely many cosets of real subtori of A.  They proved the conjecture when dim X=1.
In recent work we prove a modified version of the conjecture (it fails as stated) , giving a complete description of the topological closure of p(X), for an algebraic X. in an arbitrary compact complex torus. We prove a similar result when p:R^n->T is the covering map of a compact real torus and X a subset of R^n which  is definable in an o-minimal structure.
The result uses methods from model theory as well as some basic theory of lattices in R^n.
(Joint work with Sergei Starchenko)

Location Viernes 7 de julio a las 11:30 en el aula 520 del módulo 17


text:
hessian and complex hessian - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-22T23:16:55Z http://mathoverflow.net/feeds/question/64994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64994/hessian-and-complex-hessian hessian and complex hessian marco 2011-05-14T16:37:06Z 2011-05-14T17:21:39Z
lets assume we have a function $f : \mathbb{C}^{n} \rightarrow \mathbb{R}$ which is very nice (has all properties). Is then the complex hesse form of $f$ the same as the (real) hesse form of $f$ ? if they are not the same, then what is the connection between them ? is there any formula which brings them both in connection?