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  • This is more a matlab/math brain teaser than a question

    - by gd047
    Here is the setup. No assumptions for the values I am using. n=2; % dimension of vectors x and (square) matrix P r=2; % number of x vectors and P matrices x1 = [3;5] x2 = [9;6] x = cat(2,x1,x2) P1 = [6,11;15,-1] P2 = [2,21;-2,3] P(:,1)=P1(:) P(:,2)=P2(:) modePr = [-.4;16] TransPr=[5.9,0.1;20.2,-4.8] pred_modePr = TransPr'*modePr MixPr = TransPr.*(modePr*(pred_modePr.^(-1))') x0 = x*MixPr Then it was time to apply the following formula to get myP , where µij is MixPr. I used this code to get it: myP=zeros(n*n,r); Ptables(:,:,1)=P1; Ptables(:,:,2)=P2; for j=1:r for i = 1:r; temp = MixPr(i,j)*(Ptables(:,:,i) + ... (x(:,i)-x0(:,j))*(x(:,i)-x0(:,j))'); myP(:,j)= myP(:,j) + temp(:); end end Some brilliant guy proposed this formula as another way to produce myP for j=1:r xk1=x(:,j); PP=xk1*xk1'; PP0(:,j)=PP(:); xk1=x0(:,j); PP=xk1*xk1'; PP1(:,j)=PP(:); end myP = (P+PP0)*MixPr-PP1 I tried to formulate the equality between the two methods and seems to be this one. To make things easier, I ignored from both methods the summation of matrix P. where the first part denotes the formula that I used, while the second comes from his code snippet. Do you think this is an obvious equality? If yes, ignore all the above and just try to explain why. I could only start from the LHS, and after some algebra I think I proved it equals to the RHS. However I can't see how did he (or she) think of it in the first place.

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  • Weird Java Math ,10 ^ 1 = 11?

    - by Simon
    For an exercise I was writing a loop that turns a string into an integer without using the built in functions by multiplying each individual value by its numerical position. 75 would be 7*(10 ^ 1) + 5*(10 ^ 0), for example. However, for some reason (10 ^ 1) keeps coming back as 11. Is there some mistake I have made or an explanation for this?

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  • Math algorithm question

    - by Senica Gonzalez
    I'm not sure if this can be done without some determining factor....but wanted to see if someone knew of a way to do this. I want to create a shifting scale for numbers. Let's say I have the number 26000. I want the outcome of this algorithm to be 6500; or 25% of the original number. But if I have the number 5000, I want the outcome to be 2500; or 50% of the original number. The percentages don't have to be exact, this is just an example. I just want to have like a sine wave sort of thing. As the input number gets higher, the output number is a lower percentage of the input. Does that make sense?

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  • Closest Approach question for math/physics heads

    - by Kyle
    I'm using a Segment to Segment closest approach method which will output the closest distance between two segments of length. These objects are moving at variable speed each, so even when it succeeds I'm currently using a 10-step method and calculating the distance between 2 spheres as they move along the two segments. So, basically the length of each segment is the object's traverse in the physics step, and the radius is the objects radius. By stepping, I can tell where they collide, and if they collide (Sort of; for the MOST part.).. I get the feeling that there could be something better. While I sort of believe that the first closest approach call is required, I think that the method immediately following it is a TAD weak. Can anyone help me out? I can illustrate this if needed. Thanks alot!

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  • Math: How to sum each row of a matrix

    - by macek
    I have a 1x8 matrix of students where each student is a 4x1 matrix of scores. Something like: SCORES S [62, 91, 74, 14] T [59, 7 , 59, 21] U [44, 9 , 69, 6 ] D [4 , 32, 28, 53] E [78, 99, 53, 83] N [48, 86, 89, 60] T [56, 71, 15, 80] S [47, 67, 79, 40] Main question: Using sigma notation, or some other mathematical function, how can I get a 1x8 matrix where each student's scores are summed? # expected result TOTAL OF SCORES S [241] T [146] U [128] D [117] E [313] N [283] T [222] S [233] Sub question. To get the average, I will multiply the matrix by 1/4. Would there be a quicker way to get the final result? AVERAGE SCORE S [60.25] T [36.50] U [32.00] D [29.25] E [78.25] N [70.75] T [55.50] S [58.25] Note: I'm not looking for programming-related algorithms here. I want to know if it is possible to represent this with pure mathematical functions alone.

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  • Some math and animation

    - by Ockonal
    Hello, I have a grass texture: I use it in my 2d-game. I want to animate it by code, without any predefined animations. The grass should interact with wind. So when the wind is stronger, the grass should stoop into need side more. First version of animation I made using sinusoid function, but such animation is a bit ugly, because the base of the grass moves left/right like all another part of picture. And with sinusoid I'm not able to regulate stoop of the image. Any advices?

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  • Math/numerical formula every computer programmer should know

    - by aaa
    This is a follow-up question to What should every programmer know and Is mathematics necessary. So the question is, as a computer programmer, what is the most important/useful mathematical or numerical formula that you use? By Formula I mean anything that involves less obvious manipulations, whenever binomial coefficients or bit hacks. I work with multidimensional arrays and various matrix representations. So for me most commonly used formulas are: A(i,j,k,..) = a[i + j*Dim0 + k*Dim0*Dim1 + ... to map indexes to one dimension ( which is basic address calculation which many people do not seem to know). And triangular number T(i) = (i*i + i)/2 which is related to binomial coefficients, used to calculate address in triangular matrixes and many other things. What is your workhorse formula that you think programmer should know?

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  • iPhone SDK math - pythagorean theorem problem!

    - by Flafla2
    Just as a practice, I am working on an app that solves the famous middle school pythagorean theorem, a squared + b squared = c squared. Unfortunately, the out-coming answer has, in my eyes, nothing to do with the actual answer. Here is the code used during the "solve" action. - (IBAction)solve { int legoneint; int legtwoint; int hypotenuseint; int lonesq = legoneint * legoneint; int ltwosq = legtwoint * legtwoint; int hyposq = hypotenuseint * hypotenuseint; hyposq = lonesq + ltwosq; if ([legone.text isEqual:@""]) { legtwoint = [legtwo.text intValue]; hypotenuseint = [hypotenuse.text intValue]; answer.text = [NSString stringWithFormat:@"%d", legoneint]; self.view.backgroundColor = [UIColor blackColor]; } if ([legtwo.text isEqual:@""]) { legoneint = [legone.text intValue]; hypotenuseint = [hypotenuse.text intValue]; answer.text = [NSString stringWithFormat:@"%d", legtwoint]; self.view.backgroundColor = [UIColor blackColor]; } if ([hypotenuse.text isEqual:@""]) { legoneint = [legone.text intValue]; legtwoint = [legtwo.text intValue]; answer.text = [NSString stringWithFormat:@"%d", hypotenuseint]; self.view.backgroundColor = [UIColor blackColor]; } } By the way, legone, legtwo, and hypotenuse all represent the UITextField that corresponds to each mathematical part of the right triangle. Answer is the UILabel that tells, you guessed it, the answer. Does anyone see any flaws in the program? Thanks in advance!

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  • How to do simple math in datagridview

    - by EB
    I am new to C#. I have a datagridview with 9 columns. I am simply trying take the value in column 5 and subtract it from column 6. Then display the result in column 9. It seems simple enough.I know it's done in excel all the time. But I just cannot figure this out. Do I need to create a new class with a method called calculate columns? or does the datagridview class have something already built in that can handle this? Again, I am new to C#. Any help would be appreciated. Thank you, Oceantrain.

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  • PHP math gets crazy, need explanation, my brain is melting

    - by derei
    I know that playing with php float can give strange results if you try to add "goats + apples", but please take a look to the following case: $val = 1232.81; $p1 = 1217.16; $p2 = 15.65; $sum = $p1 + $p2; $dif = $val - $sum; echo $dif; It will give you -2.2737367544323E-13 ... yeah, ALMOST zero, but then why it doesn't say 0 ? This freaks me out big time. Please, I need some valid explanation.

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  • Step math function

    - by ekapek
    Hi I need function which returns: for any number from range = result [0.001,0.01) => 0.01 [0.01,0.1) => 0.1 [0.1,1) => 1 [1,10) => 10 [10,100) => 100 etc. My first idea was to use if, but this the worst way. Is there a simple solution?

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  • Math Looping Between Min and Max Using Mod?

    - by TheDarkIn1978
    i'm attempting to build a tiny (or perhaps not so tiny) formula that will contain numbers between a set min and max, but also loop these numbers so they are not clipped if they are outside of the range. so far, this is what i have. min1 = 10 max1 = 90 val1 = 92 //will make 12, which is what i want since it loops formula: min(max(min1,min(val1,max1)),mod(val1,max1)+min1) however, i'd like it to loop the other direction also, so that if val1 is 5, which is -5 outside of min1, it will become 85. another problem i'm running into is that max1 % max1 != max1 as i want it to, since the max is part of the range trying to be clear, here are some examples of desired output based on a range with looping min1 = 10 max1 = 90 ---------------------------------------------- val1 = 30 //within range: stays as 30 val1 = 90 //within range: stays as 90 val1 = -6 //below range: loops to becomes 84 val1 = 98 //above range: loops to becomes 18 i'd like not to resort to using a series of if/else statements, but one would be fine if it's absolutely required. is that even possible?

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  • Delphi Math: Why is 0.7<0.70?

    - by Al C
    If I have variables a, b, an c of type double, let c:=a/b, and give a and b values of 7 and 10, then c's value of 0.7 registers as being LESS THAN 0.70. On the other hand, if the variables are all type extended, then c's value of 0.7 does not register as being less than 0.70. This seems strange. What information am I missing?

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  • Math Mod Containing Numbers

    - by TheDarkIn1978
    i would like to write a simple line of code, without resorting to if statements, that would evaluate whether a number is within a certain range. i can evaluate from 0 - Max by using the modulus. 30 % 90 = 30 //great however, if the test number is greater than the maximum, using modulus will simply start it at 0 for the remaining, where as i would like to limit it to the maximum if it's past the maximum 94 % 90 = 4 //i would like answer to be 90 it becomes even more complicated, to me anyway, if i introduce a minimum for the range. for example: minimum = 10 maximum = 90 therefore, any number i evaluate should be either within range, or the minimum value if it's below range and the maximum value if it's above range -76 should be 10 2 should be 10 30 should be 30 89 should be 89 98 should be 90 23553 should be 90 is it possible to evaluate this with one line of code without using if statements?

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  • [MATH]Project an axis to a plane

    - by Robert
    how do i project the X axis to a plane if the plane is defined by three points? see pic here: http://129.25.16.135:2080/v6.5/books/usb/graphics/iconventions-local-surfaces.png The default local 1-direction is the projection of the global x-axis onto the surface. If the global x-axis is within 0.1° of being normal to the surface, the local 1-direction is the projection of the global z-axis onto the surface. let's suppose that these points are: t1 = [-0.362879991531372, 0.357021987438202, -0.373737007379532] t2 = [-0.383525013923645, 0.371621012687683, -0.383549988269806] t3 = [-0.383534014225006, 0.34629300236702, -0.38544899225235] is the vector [0.78280971952246, -0.0307519963686645, 0.411184845614438] correct answer in this case? i've calculated the angle between the surf and x (1,0,0) is ~28°. and can you give the entire procedure for the calculation, because i'm just puzzled.

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  • optimization math computation (multiplication and summing)

    - by wiso
    Suppose you want to compute the sum of the square of the differences of items: $\sum_{i=1}^{N-1} (x_i - x_{i+1})^2$, the simplest code (the input is std::vector<double> xs, the ouput sum2) is: double sum2 = 0.; double prev = xs[0]; for (vector::const_iterator i = xs.begin() + 1; i != xs.end(); ++i) { sum2 += (prev - (*i)) * (prev - (*i)); // only 1 - with compiler optimization prev = (*i); } I hope that the compiler do the optimization in the comment above. If N is the length of xs you have N-1 multiplications and 2N-3 sums (sums means + or -). Now suppose you know this variable: sum = $x_1^2 + x_N^2 + 2 sum_{i=2}^{N-1} x_i^2$ Expanding the binomial square: $sum_i^{N-1} (x_i-x_{i+1})^2 = sum - 2\sum_{i=1}^{N-1} x_i x_{i+1}$ so the code becomes: double sum2 = 0.; double prev = xs[0]; for (vector::const_iterator i = xs.begin() + 1; i != xs.end(); ++i) { sum2 += (*i) * prev; prev = (*i); } sum2 = -sum2 * 2. + sum; Here I have N multiplications and N-1 additions. In my case N is about 100. Well, compiling with g++ -O2 I got no speed up (I try calling the inlined function 2M times), why?

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  • PHP Math issue with negatives [closed]

    - by user1269625
    Possible Duplicate: PHP negatives keep adding I have this code here.... $remaining = 0; foreach($array as $value=>$row){ $remaining = $remaining + $row['remainingbalance']; } What its doing is that it is going through all the remaining balances in the array which are -51.75 and -17.85 with the code above I get -69.60 which is correct. But I am wondering how when its two negatives if they could subtract? Is that possible? I tried this $remaining = 0; foreach($clientArrayInvoice as $value=>$row){ $remaining = $remaining + abs($row['remainingbalance']); } but it gives me 69.60 without the negative. Anyone got any ideas? my goal is to take -51.75 and -17.85 and come up with -33.90 only when its a negative to do subtract. otherwise add

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  • Point inside Oriented Bounding Box?

    - by Milo
    I have an OBB2D class based on SAT. This is my point in OBB method: public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } Here is the rest of the class; the parts that pertain: public class OBB2D { private Vector2D projVec = new Vector2D(); private static Vector2D projAVec = new Vector2D(); private static Vector2D projBVec = new Vector2D(); private static Vector2D tempNormal = new Vector2D(); private Vector2D deltaVec = new Vector2D(); private ArrayList<Vector2D> collisionPoints = new ArrayList<Vector2D>(); // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(float centerx, float centery, float w, float h, float angle) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(centerx,centery,w,h,angle); } public OBB2D(float left, float top, float width, float height) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(left + (width / 2), top + (height / 2),width,height,0.0f); } public void set(float centerx,float centery,float w, float h,float angle) { float vxx = (float)Math.cos(angle); float vxy = (float)Math.sin(angle); float vyx = (float)-Math.sin(angle); float vyy = (float)Math.cos(angle); vxx *= w / 2; vxy *= (w / 2); vyx *= (h / 2); vyy *= (h / 2); corner[0].x = centerx - vxx - vyx; corner[0].y = centery - vxy - vyy; corner[1].x = centerx + vxx - vyx; corner[1].y = centery + vxy - vyy; corner[2].x = centerx + vxx + vyx; corner[2].y = centery + vxy + vyy; corner[3].x = centerx - vxx + vyx; corner[3].y = centery - vxy + vyy; this.center.x = centerx; this.center.y = centery; this.angle = angle; computeAxes(); extents.x = w / 2; extents.y = h / 2; computeBoundingRect(); } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0].x = corner[1].x - corner[0].x; axis[0].y = corner[1].y - corner[0].y; axis[1].x = corner[3].x - corner[0].x; axis[1].y = corner[3].y - corner[0].y; // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { float l = axis[a].length(); float ll = l * l; axis[a].x = axis[a].x / ll; axis[a].y = axis[a].y / ll; origin[a] = corner[0].dot(axis[a]); } } public void computeBoundingRect() { boundingRect.left = JMath.min(JMath.min(corner[0].x, corner[3].x), JMath.min(corner[1].x, corner[2].x)); boundingRect.top = JMath.min(JMath.min(corner[0].y, corner[1].y),JMath.min(corner[2].y, corner[3].y)); boundingRect.right = JMath.max(JMath.max(corner[1].x, corner[2].x), JMath.max(corner[0].x, corner[3].x)); boundingRect.bottom = JMath.max(JMath.max(corner[2].y, corner[3].y),JMath.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(rect.centerX(),rect.centerY(),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } public void moveTo(float centerx, float centery) { float cx,cy; cx = center.x; cy = center.y; deltaVec.x = centerx - cx; deltaVec.y = centery - cy; for (int c = 0; c < 4; ++c) { corner[c].x += deltaVec.x; corner[c].y += deltaVec.y; } boundingRect.left += deltaVec.x; boundingRect.top += deltaVec.y; boundingRect.right += deltaVec.x; boundingRect.bottom += deltaVec.y; this.center.x = centerx; this.center.y = centery; computeAxes(); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center.x,center.y,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center.x,center.y,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } public static float distance(float ax, float ay,float bx, float by) { if (ax < bx) return bx - ay; else return ax - by; } public Vector2D project(float ax, float ay) { projVec.x = Float.MAX_VALUE; projVec.y = Float.MIN_VALUE; for (int i = 0; i < corner.length; ++i) { float dot = Vector2D.dot(corner[i].x,corner[i].y,ax,ay); projVec.x = JMath.min(dot, projVec.x); projVec.y = JMath.max(dot, projVec.y); } return projVec; } public Vector2D getCorner(int c) { return corner[c]; } public int getNumCorners() { return corner.length; } public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } public ArrayList<Vector2D> getCollsionPoints(OBB2D b) { collisionPoints.clear(); for(int i = 0; i < corner.length; ++i) { if(b.pointInside(corner[i])) { collisionPoints.add(corner[i]); } } for(int i = 0; i < b.corner.length; ++i) { if(pointInside(b.corner[i])) { collisionPoints.add(b.corner[i]); } } return collisionPoints; } }; What could be wrong? When I getCollisionPoints for 2 OBBs I know are penetrating, it returns no points. Thanks

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  • More Fun With Math

    - by PointsToShare
    More Fun with Math   The runaway student – three different ways of solving one problem Here is a problem I read in a Russian site: A student is running away. He is moving at 1 mph. Pursuing him are a lion, a tiger and his math teacher. The lion is 40 miles behind and moving at 6 mph. The tiger is 28 miles behind and moving at 4 mph. His math teacher is 30 miles behind and moving at 5 mph. Who will catch him first? Analysis Obviously we have a set of three problems. They are all basically the same, but the details are different. The problems are of the same class. Here is a little excursion into computer science. One of the things we strive to do is to create solutions for classes of problems rather than individual problems. In your daily routine, you call it re-usability. Not all classes of problems have such solutions. If a class has a general (re-usable) solution, it is called computable. Otherwise it is unsolvable. Within unsolvable classes, we may still solve individual (some but not all) problems, albeit with different approaches to each. Luckily the vast majority of our daily problems are computable, and the 3 problems of our runaway student belong to a computable class. So, let’s solve for the catch-up time by the math teacher, after all she is the most frightening. She might even make the poor runaway solve this very problem – perish the thought! Method 1 – numerical analysis. At 30 miles and 5 mph, it’ll take her 6 hours to come to where the student was to begin with. But by then the student has advanced by 6 miles. 6 miles require 6/5 hours, but by then the student advanced by another 6/5 of a mile as well. And so on and so forth. So what are we to do? One way is to write code and iterate it until we have solved it. But this is an infinite process so we’ll end up with an infinite loop. So what to do? We’ll use the principles of numerical analysis. Any calculator – your computer included – has a limited number of digits. A double floating point number is good for about 14 digits. Nothing can be computed at a greater accuracy than that. This means that we will not iterate ad infinidum, but rather to the point where 2 consecutive iterations yield the same result. When we do financial computations, we don’t even have to go that far. We stop at the 10th of a penny.  It behooves us here to stop at a 10th of a second (100 milliseconds) and this will how we will avoid an infinite loop. Interestingly this alludes to the Zeno paradoxes of motion – in particular “Achilles and the Tortoise”. Zeno says exactly the same. To catch the tortoise, Achilles must always first come to where the tortoise was, but the tortoise keeps moving – hence Achilles will never catch the tortoise and our math teacher (or lion, or tiger) will never catch the student, or the policeman the thief. Here is my resolution to the paradox. The distance and time in each step are smaller and smaller, so the student will be caught. The only thing that is infinite is the iterative solution. The race is a convergent geometric process so the steps are diminishing, but each step in the solution takes the same amount of effort and time so with an infinite number of steps, we’ll spend an eternity solving it.  This BTW is an original thought that I have never seen before. But I digress. Let’s simply write the code to solve the problem. To make sure that it runs everywhere, I’ll do it in JavaScript. function LongCatchUpTime(D, PV, FV) // D is Distance; PV is Pursuers Velocity; FV is Fugitive’ Velocity {     var t = 0;     var T = 0;     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     t = d / pv;     while (t > 0.000001) //a 10th of a second is 1/36,000 of an hour, I used 1/100,000     {         T = T + t;         d = t * fv;         t = d / pv;     }     return T;     } By and large, the higher the Pursuer’s velocity relative to the fugitive, the faster the calculation. Solving this with the 10th of a second limit yields: 7.499999232000001 Method 2 – Geometric Series. Each step in the iteration above is smaller than the next. As you saw, we stopped iterating when the last step was small enough, small enough not to really matter.  When we have a sequence of numbers in which the ratio of each number to its predecessor is fixed we call the sequence geometric. When we are looking at the sum of sequence, we call the sequence of sums series.  Now let’s look at our student and teacher. The teacher runs 5 times faster than the student, so with each iteration the distance between them shrinks to a fifth of what it was before. This is a fixed ratio so we deal with a geometric series.  We normally designate this ratio as q and when q is less than 1 (0 < q < 1) the sum of  + … +  is  – 1) / (q – 1). When q is less than 1, it is easier to use ) / (1 - q). Now, the steps are 6 hours then 6/5 hours then 6/5*5 and so on, so q = 1/5. And the whole series is multiplied by 6. Also because q is less than 1 , 1/  diminishes to 0. So the sum is just  / (1 - q). or 1/ (1 – 1/5) = 1 / (4/5) = 5/4. This times 6 yields 7.5 hours. We can now continue with some algebra and take it back to a simpler formula. This is arduous and I am not going to do it here. Instead let’s do some simpler algebra. Method 3 – Simple Algebra. If the time to capture the fugitive is T and the fugitive travels at 1 mph, then by the time the pursuer catches him he travelled additional T miles. Time is distance divided by speed, so…. (D + T)/V = T  thus D + T = VT  and D = VT – T = (V – 1)T  and T = D/(V – 1) This “strangely” coincides with the solution we just got from the geometric sequence. This is simpler ad faster. Here is the corresponding code. function ShortCatchUpTime(D, PV, FV) {     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     return d / (pv - fv); } The code above, for both the iterative solution and the algebraic solution are actually for a larger class of problems.  In our original problem the student’s velocity (speed) is 1 mph. In the code it may be anything as long as it is less than the pursuer’s velocity. As long as PV > FV, the pursuer will catch up. Here is the really general formula: T = D / (PV – FV) Finally, let’s run the program for each of the pursuers.  It could not be worse. I know he’d rather be eaten alive than suffering through yet another math lesson. See the code run? Select  “Catch Up Time” in www.mgsltns.com/games.htm The host is running on Unix, so the link is case sensitive. That’s All Folks

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  • Kalculate = math + fun

    - by Devin A. Rychetnik
    Kalculate is a you vs. the Internet style game for math lovers. The rules are simple: answer as many math problems as you can in 90 seconds. At the end of each round, Kalculate will tally up all the scores and show you where you ranked relative to others currently playing.Tip: answering 3 questions in 10 seconds earns you a score multiplier      If you prefer to just practice and stay out of the competition, there's an offline mode that allows you to play solo.Kalculate is free (ad-supported) and can be downloaded here.

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