Tag: maths

Computational Physics Basics: How Integers are Stored

Unsigned Integers

Computers use binary representations to store various types of data. In the context of computational physics, it is important to understand how numerical values are stored. To start, let’s take a look at non-negative integer numbers. These unsigned integers can simply be translated into their binary representation. The binary number-format is similar to the all-familiar decimal format with the main difference that there are only two values for the digits, not ten. The two values are 0 and 1. Numbers are written in the same way as decimal numbers only that the place values of each digit are now powers of 2. For example, the following 4-digit numbers show the values of the first four

Binary Counter

0 0 0 1   decimal value 20 = 1
0 0 1 0   decimal value 21 = 2
0 1 0 0   decimal value 22 = 4
1 0 0 0   decimal value 23 = 8

The binary digits are called bits and in modern computers, the bits are grouped in units of 8. Each unit of 8 bits is called a byte and can contain values between 0 and 28 − 1 = 255. Of course, 255 is not a very large number and for most applications, larger numbers are needed. Most modern computer architectures support integers with 32 bits and 64 bits. Unsigned 32-bit integers range from 0 to 232 − 1 = 4, 294, 967, 295 ≈ 4.3 × 109 and unsigned 64-bit integers range from 0 to 264 − 1 = 18, 446, 744, 073, 709, 551, 615 ≈ 1.8 × 1019. It is worthwhile noting that many GPU architectures currently don’t natively support 64-bit numbers.

The computer’s processor contains registers that can store binary numbers. Thus a 64-bit processor contains 64-bit registers and has machine instructions that perform numerical operations on those registers. As an example, consider the addition operation. In binary, two numbers are added in much the same way as using long addition in decimal. Consider the addition of two 64 bit integers 7013356221863432502 + 884350303838366524. In binary, this is written as follows.

  01100001,01010100,01110010,01010011,01001111,01110010,00010001,00110110
+ 00001100,01000101,11010111,11101010,01110101,01001011,01101011,00111100
---------------------------------------------------------------------------
  01101101,10011010,01001010,00111101,11000100,10111101,01111100,01110010

The process of adding two numbers is simple. From right to left, the digits of the two numbers are added. If the result is two or more, there will be a carry-over which is added to the next digit on the left.

You could add integers of any size using this prescription but, of course, in the computer numbers are limited by the number of bits they contain. Consider the following binary addition of (264 − 1) and 1 .

  11111111,11111111,11111111,11111111,11111111,11111111,11111111,11111111
+ 00000000,00000000,00000000,00000000,00000000,00000000,00000000,00000001
---------------------------------------------------------------------------
  00000000,00000000,00000000,00000000,00000000,00000000,00000000,00000000

If you were dealing with mathematical integers, you would expect to see an extra digit 1 on the left. The computer cannot store that bit in the register containing the result but stores the extra bit in a special carry flag. In many computer languages, this unintentional overflow will go undetected and the programmer has to take care that numerical operations do not lead to unintended results.

Signed Integers

The example above shows that adding two non-zero numbers can result in 0. This can be exploited to define negative numbers. In general, given a number a, the negative  − a is defined as the number that solves the equation
a + ( − a) = 0.
Mathematically, the N-bit integers can be seen as the group of integers modulo 2N. This means that for any number a ∈ {0, …, 2N − 1} the number  − a can be defined as
 − a = 2N − a ∈ {0, …, 2N − 1}.
By convention, all numbers whose highest value binary bit is zero are considered positive. Those numbers whose highest value bit is one are considered negative. This makes the addition and subtraction of signed integers straightforward as the processor does not need to implement different algorithms for positive or negative numbers. Signed 32-bit integers range from  − 2, 147, 483, 648 to 2, 147, 483, 647, and 64-bit integers range from  − 9, 223, 372, 036, 854, 775, 808 to 9, 223, 372, 036, 854, 775, 807.

This format of storing negative numbers is called the two’s complement format. The reason for this name becomes obvious when observing how to transform a positive number to its negative.

01100001,01010100,01110010,01010011,01001111,01110010,00010001,00110110 (7013356221863432502)
10011110,10101011,10001101,10101100,10110000,10001101,11101110,11001010 (-7013356221863432502)

To invert a number, first, invert all its bits and then add 1. This simple rule of taking the two’s complement can be easily implemented in the processor’s hardware. Because of the simplicity of this prescription, and the fact that adding a negative number follows the same algorithm as adding a positive number, two’s complement is de-facto the only format used to store negative integers on modern processors.

Exercises

  1. Show that taking the two’s complement of an N-bit number a does indeed result in the negative  − a if the addition of two numbers is defined as the addition modulo 2N.
  2. Find out how integers are represented in the programming language of your choice. Does this directly reflect the representation of the underlying architecture? I will be writing another post about this topic soon.
  3. Most processors have native commands for multiplying two integers. The result of multiplying the numbers in two N-bit registers are stored in two N-bit result registers representing the high and low bits of the result. Show that the resulting 2N bits will always be enough to store the result.
  4. Show how the multiplication of two numbers can be implemented using only the bit-shift operator and conditional addition based on the bit that has been shifted out of the register. The bit-shift operator simply shifts all bits of a register to the left or right.

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String Art and Heart Curves

In my childhood, my parents would take us to the seaside for the summer holidays. One thing, other than the sun and the beach, that I distinctly remember were the nautic decorations in the cafes. One particular image was an abstract picture of a sailboat made from some nails hammered into a wooden board and pieces of string that were tied around those nails. It always fascinated me, how this simple arrangement of straight lines could create such a beautiful and dynamic image. Around the same time, in primary school, we used a similar technique in art class, with needle and thread and a piece of cardboard, to create an image of a snowflake.

It wasn’t until much later that I understood that the curves created by the thread could be described using mathematics. But while I knew that this had to do with maths somehow, during my school years I always thought that the maths was much too complicated for me to understand. Now, I have children and they have been doing similar art projects. From a teaching perspective, I am aware that these projects are intended to create an interest in maths. I think there is a missed opportunity when these line patterns are only used in primary school art lessons. The mathematics of the curves created by these lines can be accessible to secondary school students.

As an example, I want to present here the curve created by drawing straight lines in a circle. Here is the recipe:

  • Draw a large circle on a piece of paper
  • Draw a reasonably large number of equally spaced points around the circumference, about 20 to 30 should be OK
  • Choose a point as point 0 and number the points around the circle, clockwise or anticlockwise
  • Now draw a line
    • from point 1 to point 2
    • from point 2 to point 4
    • from point 3 to point 6
    • and so on, always connecting point $n$ with point $2n$.

Once the starting point has reached the opposite side, the endpoint will have reached point 0. Keep going by always incrementing the endpoint by two while moving the starting point by one.

You should end up with something like this.

As you can see the lines you have drawn create an interesting pattern. In fact, the lines trace out a curve called the cardioid or heart curve. Each line you have drawn is a tangent to the cardioid. This means that each of the lines touches the cardioid in a single point.

Let’s take a look at this cardioid curve. It is a special case of an epicycloid. This family of curves is generated by tracing a point on the circumference of one circle as it rolls around another circle. In the case of the cardioid, both circles have the same radius. The point that traces the curve is chosen to lie on the circumference of the outer circle. In the picture below you can see how the cardioid is created.

The curves in the two pictures look similar and if you were to place one image on top of the other, you would not see any difference in the curves. That seems to suggest that the two are in fact identical. But comparing images is not a mathematical proof. And surely you don’t believe this statement just because everybody on the internet claims it to be true.

If you look around, you can find several proofs that show you that the curve created by our string art is indeed the cardioid. But most of these proofs make use of trigonometry. This makes them inaccessible to anybody who is not yet comfortable with trigonometric identities. I believe that a purely geometric proof is easier to understand and also more beautiful than resorting to sine and cosine functions.

The proof

I order to prove the identity of the two curves we need to show two things. First, we need to show that our line construction intersects the cardioid in at least one point and we need to find that point. Then we need to show that the point that traces the cardioid moves in the direction of the line segment. This second part of the proof shows that the line segments are tangential to the cardioid.

Part 1: Constructing a point on the cardioid

Look at the figure below. Don’t get too confused with all the lines and angles in the picture. I will explain everything bit by bit. Remember when you were drawing the string art picture. You were drawing a line from some start point to some endpoint. In the diagram below, I have taken one of those lines and named the start point B and the endpoint B’. I have also connected the origin point 0 to the start point B. Now, B’ moves around the circle twice as fast as point B. This means that the distance from B to B’ is the same as the distance from 0 to B. In other words, if you draw a line from the centre C of the big circle to B, the figure is symmetric with respect to that line.

Now, look at the two small circles. The circle at the centre C of the large circle stays fixed. The small circle centred at C’ on the connection between C and B is the circle that rolls along the circumference and creates the cardioid. The line B-B’ intersects this outer circle at a point P. We need to show that this intersection point P is the same as the point that stays fixed on the outer circle as it rolls along. We can prove this by showing that the angle C-C’-P is the same as the angle that C-C’ makes with the vertical through the central point C.

Take the intersection between the line A-B and the rolling circle and call it P’. You can immediately see that the triangles A-B-C and P’-B-C’ are similar. This means that the line P’-C’ is vertical, just like the line A-C. It follows that the angle C-C’-P’ is the same as the angle that C-C’ makes with the vertical because they are opposite angles. But because of the symmetry of the construction, this angle is also the same as C-C’-P.

With this, we have shown that the intersection point P is, in fact, a point on the cardioid curve.

Part 2: Direction of motion of P

The motion of the point P, as the outer small circle rolls around the inner circle, is made up of two contributions. The first is the motion of the centre C’ orbiting around the inner circle. The second is the motion due to the rolling circle spinning around its own centre. We need to look at the direction and magnitude of each of these velocities.
Let’s say that the radius of the small circles is 1 and the angular velocity of the rotation of the outer circle around the inner is also 1. The exact values are not important because we only need to compare the two contributions of the velocity of point P. We don’t need to find its absolute value.

Orbital Velocity

The orbital velocity is oriented perpendicular to the connection C-B which also is parallel to the tangent to the circle through B. This means that naturally, the angle between this velocity and the line B-P is the same as the angle between the tangent and B-P. To compare the magnitudes we have to remember some equation from physics. The speed of a point rotating around a centre is the product of the distance to the centre and the angular velocity. Let’s call the radius of the small circle $R$ and the angular velocity of the rotation of the outer circle around the inner circle $\omega$ (that’s the lowercase Greek letter omega). The distance of the outer centre to the centre of rotation is $2R$ and this makes the speed of the orbital rotation equal to
$$v_{\text{orbit}} = 2R\omega.$$

Now let’s look at the velocity of the spin rotation. The point P rotates around the centre of rotation C’. The radius is $R$ but the angular velocity is $2\omega$. To see this, look at the vertical line C’-P’. This line stays vertical at all times. In part 1 of the proof we already saw that that the angle C-C’-P’ is the same as the angle that C-C’ makes with the vertical, so as C’ moves around C by some angle, C-C’-P’ will increase by the same amount. But so will C-C’-P. This means that the angle P’-C-P increases twice as fast as the orbital angle. The resulting speed of the spin motion is
$$v_{\text{spin}} = R\times 2\omega.$$
We can see that the magnitude of the velocity of the spin motion is equal to that of the orbital motion. We also saw that the line B-P halves the angle between those two velocities. This means that the sum of these two velocities must lie in the direction of B-P. In other words, B-P is a tangent to the motion of the point P which is what we needed to prove.

Summary

I admit that this proof is somewhat lengthy but I do think that it gives some interesting insight into why the pattern of lines created by our initial construction generates the cardioid curve. All the figures for this post were created with Geogebra and are available online.

 

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Beware of square roots and exponentials!

In the last post I asked you the question, what is wrong with the following proof.

As some of you noticed, the answer lies in the fact that the identity

is only true for positive real numbers x and y. In general one has to be careful with the identity

(xy)a = xaya

for non-integer a and non-positive x and y.

I posted this question because, although it ought to be, it is not always taught to students when complex numbers are introduced. Once students are taught complex numbers they feel that, now that they are free to take the square root of any real number, they don’t have to pay attention anymore to what they are doing. They apply the rules blindly, resting in false comfort that everything is now fine.

Another pitfall with complex numbers, that even educated people are not always aware of, is the fact that

(x^y)^z \ne x^{yz}

if y and z are complex numbers, even for positive, real x. If it were we, could prove the following (here I assume that you are familiar with the complex exponential function).

Let’s start with the following identity

e^{1+2i\pi n}=e\times e^{2i\pi n} = e\times 1 = e

This makes

(e1 + 2iπn)2iπn = (e)2iπn = 1

But, if the above identity were true, the last line would be equal to

e^{2i\pi n - 4\pi^2 n^2} = e^{2i\pi n} e^{-4\pi^2 n^2}= e^{- 4\pi^2 n^2}

Which, in general, is a positive real number not equal to 1.

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Another proof that -1=1

Here is a little maths teaser for you. Since I was a student, I always loved those “proofs” that zero equals one. Of course, most of the time, this was achieved by sneakily dividing by zero somewhere along the way.

But yesterday I came across a proof that used a different, slightly more subtle trick and uses complex numbers. I apologise to any reader not familiar with complex numbers. Anyone interested can find a quick introduction here.

Enough introduction, here is the “proof”:

Looks OK, but it can’t be right of course. So where is the error in this equation? Can you find out?

Update:
You can find the answer here.

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