User Tag List

First 89101112 Last

Results 91 to 100 of 121

  1. #91

    Default

    Quote Originally Posted by CaptainChick View Post
    Is pi a finite number/quantity?
    Yes. It is irrational, but finite.

    Quote Originally Posted by CaptainChick View Post
    Also, is a discrete, actual circle even real?!?!?!?

    I have issues with pi.

    Is a circle a theoretical construct?

    Like, in reality, what does the circumference, (as in the linear perimeter) of a circle even look like?

    Sorry, if I am asking stupid questions. :/
    B. Fuller also didn't like the ideas of perfect circles in nature.

    I doubt any real examples exist. It's just an abstraction (a very usefull one). The closest things I can think of are the orbits of physical systems in uniform circular motion (like in centrifuges).

    Interestingly enough, pi shows up in statistics often through spectral analasys because of the close relationship between Gaussians and Sinusoids through the Singular Value Decomposition.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  2. #92
    Banned
    Join Date
    Jul 2008
    MBTI
    type
    Posts
    9,100

    Default

    Quote Originally Posted by ygolo View Post
    B. Fuller also didn't like the ideas of perfect circles in nature.

    I doubt any real examples exist. It's just an abstraction (a very usefull one). The closest things I can think of are the orbits of physical systems in uniform circular motion (like in centrifuges).
    Don't those theoretically always end up being ellipses?

  3. #93

    Default

    Quote Originally Posted by Jack Flak View Post
    Don't those theoretically always end up being ellipses?
    They are probably subtley different from idealized shapes (which elipses are too).

    Circular motion is, however, a close enough for many calculations.

    There is a reference frame one can create (often) where the coordinates are referenced to the unit vectors that are normal, tangential, and axial to the direction of motion.

    Of course, this is often not an inertial refrence frame, and many times you can decompose the path into tiny (infinitesimal) "arcs" of uniform circular motion. The tangential vector comes in the direction of "rotation." The normal vector in the direction of force, and the axial vector a cross-product of the other two.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  4. #94

    Default Rational numbers

    I want to expound on some extremely basic properties rational numbers that I find fascinating.

    First, for the uninitiated, rational numbers are the numbers that result from dividing an integer by non-zero integer. I will assume you understand the properties of the Integers.

    Two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers) are considered "equal" if and only if p*s=r*q in the integer sense of equality.

    You multiply two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers) to get a rational number l=(p*r)/(q*s). When you add those two same rational numbers, you get (p*s+r*q)/(2*q*s).

    Also note, for a rational number of the form, n=p/1(where p is an integer), we can shord hand the rational number to just be the integer p.

    For those of you who wonder if this is a valid definition of equality:
    1. A rational number is equal to itself. If m is a rational number given by dividing r by s. r*s=r*s. (reflexive property)
    2. Given two rational numbers m=p/q, and n=r/s (where p, and r are integers, and s and q are non-zero integers). If m=n, p*s=r*q. This means r*q=s*p*s, which means m=n. (symmetric property)
    3. Given three rational numbers n=p/q, m=r/s, and l=t/u (where p, r, and t are integers, and s, q and u are non-zero integers)... and given that n=m and m=l. We have p*s=r*q and r*u=s*t. This means p*s*r*u=r*q*s*t. This further means that p*u*r*s=q*t*r*s. Dividing by r*s, p*u=q*t. This means n=l. (transitive property)


    Given two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers) n is less than or equal to m if and only if p*s is less than or equal to r*q in the integer sense of the inequality.

    For those of you uncomfortable with this definition of the inequality, we'll check the properties.
    1. Given two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers). If n is less than or equal to m, and m is less than or equal to n. p*s is less than or equal to r*q, and r*q is less than or equal to p*s in the integer sense. This means in the integer sense, p*s=r*q. This then means n=m. (anti-symmetry).
    2. Given three rational numbers n=p/q, m=r/s, and l=t/u (where p, r, and t are integers, and s, q and u are non-zero integers)... and given that n is less than or equal to m and m is less than or equal to l. We have p*s is less than or equal to r*q and r*u less than or equal to s*t. This means p*s*r*u less than or equal to r*q*s*t. This further means that p*u*r is less than or equal to s=q*t*r*s. Dividing by r*s, p*u is less than or equal to q*t. This means n=l. (transitivity)
    3. Given two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers), either n*s is less than or equal to r*q or r*q is less than or equal to n*s or both. This means n is less than or equal to m or m is less than or equal to n


    We say n is greater than or equal to m is m if an only if less than or equal to n. We say that m is less than(<) n if and only if m is less than or equal to n and it is not true that m=n. We say that n is greater than (>) m if and only if it is not true that n is less than or equal to n.

    OK, now that you have been initiated to rational numbers, I'd like to point out some interesting properties.

    First, between every pair of rational numbers, there is the average of the two rational numbers between them..

    More formally, consider two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers), where n<m. This means, p*s<r*q. Consider another rational number, l=(p*s+r*q)/(2*q*s). 2*q*s*p < q*p*s+q*r*q=q*(p*s+r*q), This means n<l. Also, s*(p*s+r*q)=s*p*s+s*r*q<2*s*r*q=2*q*s*r. This means l<m. So for any two rational numbers, there is a rational number (namely the average) between them.

    Rational numbers have a "natural" form (my term, useful for the ensuing discussion). Say n=p/q. If there exists an integer, r, greater than 1, such that r divides p evenly, and r divides s evenly, then p/q is in an “unnatural” form of n. We can divide both p and q by r and check to see if the result for a common divisor. Once n=p_0/q_0, where p_0 and q_0 have no integer divisors in common that are greater than 1, well call it the “natural” form.

    So with all these numbers around, one may think that every point along a number line is covered, but this is far from the case.

    First, let's see if we can find a rational number n, such that n^2=2. Let us suppose we have such a rational number, n, in natural form, n=p/q. Then p^2/q^2=2. This means p^2=2*q^2. So p^2 is even, meaning p is even. So p=2*r (r is an integer). This means (2r)^2=4*r^2=2*q^2. So 2*r^2=q^2. This implies that q^2 is even, and therefore q is even. However, this means that n was not in natural form, no matter what we choose from p and q. So there is no rational solution to n^2=2.

    Let's look now at the two sets of rational numbers defined by A={all rational numbers, n, such that n^2<2}, B={all rational numbers, n, such that n^2>2}.

    Let us now define a function that returns a rational result for any rational input, n>0.

    m=n+(2-n^2)/(n+2). Notice:m>n if and only n^2<2. Similarly, m<n if and only if n^2>2.

    We can re-write m like this: m=(2*n+2)/(n+2). So now m^2-2=2*(n^2-2)/(n+2)^2. Notice now that, m^2-2<0 (IOW m^2<2) if and only if n^2<2, and that m^2>2 if and only if n^2>2.

    So now for every element, n, in A, we have a greater element, m in A. This means there is no greatest element in A.

    Similarly for every element, n, in B, we have a lesser element, m in B. This means there is no least element in B.

    Trippy huh?

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  5. #95

    Default

    Quote Originally Posted by ygolo View Post
    Consider the following 4-dimensional hypercube with MBTI types on each node:

    MBTIc Math thread-mbti_hypercube-jpg

    A Hamiltonian Path is (informally) a way to visit each node/vertex in a graph without picking up your pencil.

    Can you find all Hamiltonian Paths on the MBTI hyper-cube?

    One such path was used in my post on MBTI superlatives.

    How many different such Paths are there?
    Just want to say, I solved this (again?) recently. I wrote a brute force C++ program to solve it(may post it later).

    Major props to whoever else figures it out. This is MUCH harder that the stuff in the puzzles thread.

    There is a particular search engine that'll give you approximate answers. But try to get exact answers...and more importantly, a solution with explanation.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  6. #96
    Glowy Goopy Goodness The_Liquid_Laser's Avatar
    Join Date
    Jul 2007
    MBTI
    ENTP
    Posts
    3,377

    Default

    Quote Originally Posted by ygolo View Post
    You multiply two rational numbers, n=p/q, and m=r/s, (where p, and r are integers, and s and q are non-zero integers) to get a rational number l=(p*r)/(q*s). When you add those two same rational numbers, you get (p*s+r*q)/(2*q*s).
    I don't believe you should have a 2 in the denominator here. p/q + r/s = (p*s+r*q)/(q*s)
    I mention this because you also use this expression later in your post with the 2 still in it. Perhaps you were just seeing if I was paying attention.
    My wife and I made a game to teach kids about nutrition. Please try our game and vote for us to win. (Voting period: July 14 - August 14)
    http://www.revoltingvegetables.com

  7. #97

    Default

    Quote Originally Posted by The_Liquid_Laser View Post
    I don't believe you should have a 2 in the denominator here. p/q + r/s = (p*s+r*q)/(q*s)
    I mention this because you also use this expression later in your post with the 2 still in it. Perhaps you were just seeing if I was paying attention.
    haha, yeah, where did that 2 come from? It's good to know someone is paying attention.

    Probably a copy and paste error from when I was talking about the everage od two rational numbers.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  8. #98
    Senior Member laughingebony's Avatar
    Join Date
    Jun 2009
    MBTI
    INTP
    Posts
    236

    Default

    Yesterday, as I was thinking about nothing even remotely related to logic or math, it hit me:

    "Mathematical induction" is a misnomer.

    Mathematical induction constructs an argument of the following form:

    If A, then B.
    A.
    Therefore, B.

    More specifically...

    If one case is true, all cases after it are true.
    One case is true.
    Therefore, all cases after it are true.

    This is deductive.

    It then proves the two premises deductively within the system of mathematics. There is nothing inductive about mathematical induction.

  9. #99

    Default

    Here is my code for the Hamiltonian path problem:
    (Note the problem is equivalent to finding the number of n-bit Gray Codes.)
    PHP Code:
    #include <iostream>
    using namespace std;
    #include <queue>

    // This program calculatges the Gray Codes
    // for a limited number of bits (<=4)
    class GrayCode;
    class 
    CodeQueue;

    void fail(const char* const str);

    class 
    GrayCode{
    public:
      
    enum{NUMBITS=5};
      
    enum{MAXCODE=0x20};
      
    //enum{NUMBITS=4};
      //enum{MAXCODE=0x10};
      //enum{NUMBITS=3};
      //enum{MAXCODE=0x8};
      //enum{NUMBITS=2};
      //enum{MAXCODE=0x4};
      
    GrayCode();
      
    GrayCode(const GrayCodecode);
      
    bool addCode(int code);
      
    int getLength()const {return length;}
      
    bool getLast(intlast) const;
      
    void print(ostream stream) const;
    private:
      
    int length;
      
    int bitField[MAXCODE]; //will keep number of bits small
      
    bool used[MAXCODE]; // used number in code
    };

    class 
    CodeQueue{
    public:
      
    CodeQueue(){}
      
    bool enqueue(GrayCode code){_q.push(code); return true;}
      
    bool dequeue(GrayCode*& code){code=_q.front(); _q.pop(); return true;}
      
    bool empty(){return _q.empty(); }
      
    queue<GrayCode*>::size_type getLength(){ return _q.size();}
      static 
    void processQueue(CodeQueue &workQCodeQueue &doneQ);
    private:
      
    queue<GrayCode*> _q;
    };

    int main(){
      
    cout<<"Calculating Gray Codes"<<endl;
      
    cout<<"Maxes: "<<GrayCode::MAXCODE<<endl;

      
    CodeQueue workQdoneQ;

      for(
    int i=0i<GrayCode::MAXCODE; ++i){
        
    GrayCode start = new GrayCode();
        if(!
    start->addCode(i)){fail("addCode failed");}
        if(!
    workQ.enqueue(start)){fail("enqueue failed");}
      }

      
    cout<<"There are "<<workQ.getLength()<<" "<<GrayCode::NUMBITS;
      
    cout<<"-bit Start Codes"<<endl;

      
    CodeQueue::processQueue(workQ,doneQ);

      
    cout<<"There are "<<doneQ.getLength()<<" "<<GrayCode::NUMBITS;
      
    cout<<"-bit Gray Codes"<<endl;

      
    //while(!doneQ.empty()){
      //GrayCode *toPrint;
      //if(!doneQ.dequeue(toPrint)){fail("enqueue failed");}
      //toPrint->print(cout);
      //}

      
    return 0;
    }

    void fail(const char* const str){
      
    cerr << str << endl;
      exit(
    1);
    }

    GrayCode::GrayCode():length(0){
      for(
    int i=0i<MAXCODE; ++i){used[i]=false;}
    }

    GrayCode::GrayCode(const GrayCodecode):length(code.length){
      for(
    int i=0i<MAXCODE; ++i){used[i]=code.used[i];}
      for(
    int i=0i<length; ++i){bitField[i]=code.bitField[i];}  
    }

    bool GrayCode::addCode(int code){
      if(
    length>=MAXCODE){return false;}
      if(
    used[code]){return false;}
      
    bitField[length++]=code;
      
    used[code]=true;
      return 
    true;
    }

    bool GrayCode::getLast(intlast) const {
      if(
    0==length){return false;}
      
    last=bitField[length-1];
      return 
    true;
    }

    void GrayCode::print(ostream stream) const{
      for(
    int i=0i<length; ++i){stream<<bitField[i]<<" ";}
      
    stream<<endl;    
    }

    void CodeQueue::processQueue(CodeQueue &workQCodeQueue &doneQ){
      while(!
    workQ.empty()){
        
    GrayCodetoWork;
        if(!
    workQ.dequeue(toWork)){fail("dequeue Failed");}
        
    cerr<<".";
        if(
    toWork->getLength()==GrayCode::MAXCODE){
          if(!
    doneQ.enqueue(toWork)){fail("enqueue failed");} 
        } else {
          
    GrayCodetoAdd =new GrayCode(*toWork);
          
    int last;
          if(!
    toAdd->getLast(last)){fail("getLast Failed");}
          for(
    int i=0,bit=1i<GrayCode::NUMBITS; ++ibit<<=1){
            if(
    toAdd->addCode(last^bit)){
              
    workQ.enqueue(toAdd);
              
    toAdd =new GrayCode(*toWork);
            }
          }
          
    delete toWork;
        }
      }
      
    cerr<<endl;

    Obviously NOT php, I just liked the formating.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

  10. #100

    Default

    OK, so nightning, antireconciler, and I have been having an ongoing discussion about science and its nature.

    We also spent (are spending) a lot of time on statistics. From this I thought of several challenges for people who are interested in understanding the statistical methods commonly used in science.

    The actual math for these ought to be straightforward. The idea is to really examine the assumptions behind the methods.

    I will try to find some good links illustrating the methods needed to address these challenge problems.

    Quote Originally Posted by ygolo View Post
    Audio Blog 4

    This is part of the on-going discussion about science we are having. I apologize for the cross-posting that I am doing, but I feel it appropriate.

    The challenges as promised in the audio blog:
    1. Prove that the sum of squares about the sample mean, divided by the population variance is a chi-square-distribution with n-1 degrees of freedom.
    2. Prove that the F-score used for a one-way ANOVA does indeed follow an F-distribution if the null hypothesis is true.
    3. Prove that the following tests will follow the t-distribution if the null hypothesis is true:
      1. independent one-sample t-test
      2. paired two-sample t-test
      3. unpaired two-sample t-test w/ arbitrary sample sizes and variances
    4. Prove that the F-score used for a one-way ANOVA with only two-levels for a factor for comparison is the square of the t-score for a 2 sample t-test using the same data.
    5. Prove or disprove that in a two-way ANOVA, the Total Sum of Squares (TSS) is less than the sum of the Row Sum of Squares (RSS) and the Column Sum of Squares (CSS). In other words, prove of disprove that the Error Sum of Squares (ESS) = TSS-RSS-CSS is always positive.
    6. Prove or disprove that in a two-way ANOVA, the number of degrees of freedom for the TSS is equal to the sum of the number of degrees of freedom of the RSS, CSS, and ESS.
    7. Prove or Disprove that the one-way ANOVA is a special case of the two-way ANOVA with the number of rows or columns reduced to 1 (Of course only one of the F-scores will make sense. The question is this F-score is the same as the F-score one gets for the one-way ANOVA)
    8. Prove or Disprove that the one-way ANOVA is a special case of the two-way ANOVA when one of the factors is ignored. (This time there is the same amount of "rows" and "columns", except you run a one-way ANOVA on the data).
    Links:
    Analysis of variance - Wikipedia, the free encyclopedia
    F-test - Wikipedia, the free encyclopedia
    Student's t-test - Wikipedia, the free encyclopedia
    Stats: Two-Way ANOVA

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

Similar Threads

  1. The Every-Member-Of-MBTIc Appreciation Thread
    By spirilis in forum The Fluff Zone
    Replies: 18
    Last Post: 10-18-2008, 10:21 PM
  2. The Everyone-Who's-Not-a-Member-of-MBTIc Appreciation Thread
    By ThatsWhatHeSaid in forum The Fluff Zone
    Replies: 5
    Last Post: 10-17-2008, 02:35 PM
  3. MBTIc Appreciation Thread
    By ThatsWhatHeSaid in forum The Fluff Zone
    Replies: 49
    Last Post: 10-16-2008, 02:19 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
Single Sign On provided by vBSSO