Tag: maths

Computational Physics Basics: Polynomial Interpolation

The piecewise constant interpolation and the linear interpolation seen in the previous post can be understood as special cases of a more general interpolation method. Piecewise constant interpolation constructs a polynomial of order 0 that passes through a single point. Linear interpolation constructs a polynomial of order 1 that passes through 2 points. We can generalise this idea to construct a polynomial of order \(n-1\) that passes through \(n\) points, where \(n\) is 1 or greater. The idea is that a higher-order polynomial will be better at approximating the exact function. We will see later that this idea is only justified in certain cases and that higher-order interpolations can actually increase the error when not applied with care.

Existence

The first question that arises is the following. Given a set of \(n\) points, is there always a polynomial of order \(n-1\) that passes through these points, or are there multiple polynomials with that quality? The first question can be answered simply by constructing a polynomial. The simplest way to do this is to construct the Lagrange Polynomial. Assume we are given a set of points, \[
(x_1, y_1), (x_2, y_2) \ldots (x_n, y_n),
\]
where all the \(x\)’s are different, i.e. \(x_i \ne x_j\) if \(i \ne j\). Then we observe that the fraction \[
\frac{x – x_j}{x_i – x_j}
\]
is zero when \(x = x_j\) and one when \(x = x_i\). Next, let’s choose an index \(i\) and multiply these fractions together for all \(j\) that are different to \(i\), \[
a_i(x) = \frac{x – x_1}{x_i – x_1}\times \ldots\times\frac{x – x_{i-1}}{x_i – x_{i-1}}
\frac{x – x_{i+1}}{x_i – x_{i+1}}\times \ldots\times\frac{x – x_n}{x_i – x_n}.
\]
This product can be written a bit more concisely as \[
a_i(x) = \prod_{\stackrel{j=1}{j\ne i}}^n \frac{x – x_j}{x_i – x_j}.
\]
You can see that the \(a_i\) are polynomials of order \(n-1\). Now, if \(x = x_i\) all the factors in the product are 1 which means that \(a_i(x_i) = 1\). On the other hand, if \(x\) is any of the other \(x_j\) then one of the factors will be zero and \(a_i(x_j) = 0\) for any \(j \ne i\). Thus, if we take the product \(a_i(x) y_i\) we have a polynomial that passes through the point \((x_i, y_i)\) but is zero at all the other \(x_j\). The final step is to add up all these separate polynomials to construct the Lagrange Polynomial, \[
p(x) = a_1(x)y_1 + \ldots a_n(x)y_n = \sum_{i=1}^n a_i(x)y_i.
\]
By construction, this polynomial of order \(n-1\) passes through all the points \((x_i, y_i)\).

Uniqueness

The next question is if there are other polynomials that pass through all the points, or is the Lagrange Polynomial the only one? The answer is that there is exactly one polynomial of order \(n\) that passes through \(n\) given points. This follows directly from the fundamental theorem of algebra. Imagine we have two order \(n-1\) polynomials, \(p_1\) and \(p_2\), that both pass through our \(n\) points. Then the difference, \[
d(x) = p_1(x) – p_2(x),
\]
will also be an order \(n-1\) degree polynomial. But \(d\) also has \(n\) roots because \(d(x_i) = 0\) for all \(i\). But the fundamental theorem of algebra asserts that a polynomial of degree \(n\) can have at most \(n\) real roots unless it is identically zero. In our case \(d\) is of order \(n-1\) and should only have \(n-1\) roots. The fact that it has \(n\) roots means that \(d \equiv 0\). This in turn means that \(p_1 = p_2\) must be the same polynomial.

Approximation Error and Runge’s Phenomenon

One would expect that the higher order interpolations will reduce the error of the approximation and that it would always be best to use the highest possible order. One can find the upper bounds of the error using a similar approach that I used in the previous post on linear interpolation. I will not show the proof here, because it is a bit more tedious and doesn’t give any deeper insights. Given a function \(f(x)\) over an interval \(a\le x \le b\) and sampled at \(n+1\) equidistant points \(x_i = a + hi\), with \(i=0, \ldots , n+1\) and \(h = (b-a)/n\), then the order \(n\) Lagrange polynomial that passes through the points will have an error given by the following formula. \[
\left|R_n(x)\right| \leq \frac{h^{n+1}}{4(n+1)} \left|f^{(n+1)}(x)\right|_{\mathrm{max}}
\]
Here \(f^{(n+1)}(x)\) means the \((n+1)\)th derivative of the the function \(f\) and the \(\left|.\right|_{\mathrm{max}}\) means the maximum value over the interval between \(a\) and \(b\). As expected, the error is proportional to \(h^{n+1}\). At first sight, this implies that increasing the number of points, and thus reducing \(h\) while at the same time increasing \(n\) will reduce the error. The problem arises, however, for some functions \(f\) whose \(n\)-th derivatives grow with \(n\). The example put forward by Runge is the function \[
f(x) = \frac{1}{1+25x^2}.
\]

Interpolation of Runge’s function using higher-order polynomials.

The figure above shows the Lagrange polynomials approximating Runge’s function over the interval from -1 to 1 for some orders. You can immediately see that the approximations tend to improve in the central part as the order increases. But near the outermost points, the Lagrange polynomials oscillate more and more wildly as the number of points is increased. The conclusion is that one has to be careful when increasing the interpolation order because spurious oscillations may actually degrade the approximation.

Piecewise Polynomial Interpolation

Does this mean we are stuck and that moving to higher orders is generally bad? No, we can make use of higher-order interpolations but we have to be careful. Note, that the polynomial interpolation does get better in the central region when we decrease the spacing between the points. When we used piecewise linear of constant interpolation, we chose the points that were used for the interpolation based on where we wanted to interpolate the function. In the same way, we can choose the points through which we construct the polynomial so that they are always symmetric around \(x\). Some plots of this piecewise polynomial interpolation are shown in the plot below.

Piecewise Lagrange interpolation with 20 points for orders 0, 1, 2, and 3.

Let’s analyse the error of these approximations. Using an array with \(N\) points on Runge’s function equally spaced between -2 and 2. \(N\) was varied between 10 and 10,000. For each \(N\), the centred polynomial interpolation of orders 0, 1, 2, and 3 was created. Finally, the maximum error of the interpolation and the exact function over the interval -1 and 1 are determined.

Scaling of the maximum error of the Lagrange interpolation with the number of points for increasing order.

The plot above shows the double-logarithmic dependence of the error against the number of points for each order interpolation. The slope of each curve corresponds to the order of the interpolation. For the piecewise constant interpolation, an increase in the number of points by 3 orders of magnitude also corresponds to a reduction of the error by three orders of magnitude. This indicates that the error is first order in this case. For the highest order interpolation and 10,000 points, the error reaches the rounding error of double precision.

Discontinuities and Differentiability

As seen in the previous section, for many cases the piecewise polynomial interpolation can provide a good approximation to the underlying function. However, in some cases, we need to use the first or second derivative of our interpolation. In these cases, the Lagrange formula is not ideal. To see this, the following image shows the interpolation error, again for Runge’s function, using order 2 and 3 polynomials and 20 points.

Error in the Lagrange interpolation of Runge’s function for orders 2 and 3.

One can see that the error in the order 2 approximation has discontinuities and the error in the order 3 approximation has discontinuities of the derivative. For odd-order interpolations, the points that are used for the interpolation change when \(x\) moves from an interval \([x_{i-1},x_i]\) to an interval \([x_i, x_{i+1}]\). Because both interpolations are the same at the point \(x_i\) itself, the interpolation is continuous but the derivative, in general, is not. For even-order interpolations, the stencil changes halfway between the points, which means that the function is discontinuous there. I will address this problem in a future post.

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Computational Physics Basics: Piecewise and Linear Interpolation

One of the main challenges of computational physics is the problem of representing continuous functions in time and space using the finite resources supplied by the computer. A mathematical function of one or more continuous variables naturally has an infinite number of degrees of freedom. These need to be reduced in some manner to be stored in the finite memory available. Maybe the most intuitive way of achieving this goal is by sampling the function at a discrete set of points. We can store the values of the function as a lookup table in memory. It is then straightforward to retrieve the values at the sampling points. However, in many cases, the function values at arbitrary points between the sampling points are needed. It is then necessary to interpolate the function from the given data.

Apart from the interpolation problem, the pointwise discretisation of a function raises another problem. In some cases, the domain over which the function is required is not known in advance. The computer only stores a finite set of points and these points can cover only a finite domain. Extrapolation can be used if the asymptotic behaviour of the function is known. Also, clever spacing of the sample points or transformations of the domain can aid in improving the accuracy of the interpolated and extrapolated function values.

In this post, I will be talking about the interpolation of functions in a single variable. Functions with a higher-dimensional domain will be the subject of a future post.

Functions of a single variable

A function of a single variable, \(f(x)\), can be discretised by specifying the function values at sample locations \(x_i\), where \(i=1 \ldots N\). For now, we don’t require these locations to be evenly spaced but I will assume that they are sorted. This means that \(x_i < x_{i+1}\) for all \(i\). Let’s define the function values, \(y_i\), as \[
y_i = f(x_i).
\]
The intuitive idea behind this discretisation is that the function values can be thought of as a number of measurements. The \(y_i\) provide incomplete information about the function. To reconstruct the function over a continuous domain an interpolation scheme needs to be specified.

Piecewise Constant Interpolation

The simplest interpolation scheme is the piecewise constant interpolation, also known as the nearest neighbour interpolation. Given a location \(x\) the goal is to find a value of \(i\) such that \[
|x-x_i| \le |x-x_j| \quad \text{for all} \quad j\ne i.
\]
In other words, \(x_i\) is the sample location that is closest to \(x\) when compared to the other sample locations. Then, define the interpolation function \(p_0\) as \[
p_0(x) = f(x_i)
\]
with \(x_i\) as defined above. The value of the interpolation is simply the value of the sampled function at the sample point closest to \(x\).

Piecewise constant interpolation of a function (left) and the error (right)

The left plot in the figure above shows some smooth function in black and a number of sample points. The case where 10 sample points are taken is shown by the diamonds and the case for 20 sample points is shown by the circles. Also shown are the nearest neighbour interpolations for these two cases. The red curve shows the interpolated function for 10 samples and the blue curve is for the case of 20 samples. The right plot in the figure shows the difference between the original function and the interpolations. Again, the red curve is for the case of 10 samples and the blue curve is for the case of 20 samples. We can see that the piecewise constant interpolation is crude and the errors are quite large.

As expected, the error is smaller when the number of samples is increased. To analyse exactly how big the error is, consider the residual for the zero-order interpolation \[
R_0(x) = f(x) – p_0(x) = f(x) – f(x_i).
\]
The first step to analyse the magnitude of the residual is to perform a Taylor expansion of the residual around the point \(x_i\). We only need the zero order term. Using Taylor’s Theorem and the Cauchy form of the remainder, one can write \[
R_0(x) = \left[ f(x_i) + f'(\xi_c)(x – x_i)\right] – f(x_i).
\]
The term in the brackets is the Taylor expansion of \(f(x)\), and \(\xi_c\) is some value that lies between \(x_i\) and \(x\) and depends on the value of \(x\). Let’s define the distance between two samples with \(h=x_{i+1}-x_i\). Assume for the moment that all samples are equidistant. It is not difficult to generalise the arguments for the case when the support points are not equidistant. This means, the maximum value of \(x – x_i\) is half of the distance between two samples, i.e. \[
x – x_i \le \frac{h}{2}.
\]
It os also clear that \(f'(\xi_c) \le |f'(x)|_{\mathrm{max}}\), where the maximum is over the interval \(|x-x_i| \le h/2\). The final result for an estimate of the residual error is \[
|R_0(x)| \le\frac{h}{2} |f'(x)|_{\mathrm{max}}
\]

Linear Interpolation

As we saw above, the piecewise interpolation is easy to implement but the errors can be quite large. Most of the time, linear interpolation is a much better alternative. For functions of a single argument, as we are considering here, the computational expense is not much higher than the piecewise interpolation but the resulting accuracy is much better. Given a location \(x\), first find \(i\) such that \[
x_i \le x < x_{i+1}.
\]
Then the linear interpolation function \(p_1\) can be defined as \[
p_1(x) = \frac{x_{i+1} – x}{x_{i+1} – x_i} f(x_i)
+ \frac{x – x_i}{x_{i+1} – x_i} f(x_{i+1}).
\]
The function \(p_1\) at a point \(x\) can be viewed as a weighted average of the original function values at the neighbouring points \(x_i\) and \(x_{i+1}\). It can be easily seen that \(p(x_i) = f(x_i)\) for all \(i\), i.e. the interpolation goes through the sample points exactly.

Linear interpolation of a function (left) and the error (right)

The left plot in the figure above shows the same function \(f(x)\) as the figure in the previous section but now together with the linear interpolations for 10 samples (red curve) and 20 samples (blue curve). One can immediately see that the linear interpolation resembles the original function much more closely. The right plot shows the error for the two interpolations. The error is much smaller when compared to the error for the piecewise interpolation. For the 10 sample interpolation, the maximum absolute error of the linear interpolation is about 0.45 compared to a value of over 1.5 for the nearest neighbour interpolation. What’s more, going from 10 to 20 samples improves the error substantially.

One can again try to quantify the error of the linear approximation using Taylor’s Theorem. The first step is to use the Mean Value Theorem that states that there is a point \(x_c\) between \(x_i\) and \(x_{i+1}\) that satisfies \[
f'(x_c) = \frac{ f(x_{i+1}) – f(x_i) }{ x_{i+1} – x_i }.
\]
Consider now the error of the linear approximation, \[
R_1(x) = f(x) – p_1(x) = f(x) – \left[\frac{x_{i+1} – x}{x_{i+1} – x_i} f(x_i)
+ \frac{x – x_i}{x_{i+1} – x_i} f(x_{i+1})\right].
\]
The derivative of the error is \[
R’_1(x) = f'(x) – \frac{ f(x_{i+1}) – f(x_i) }{ x_{i+1} – x_i }.
\]
The Mean Value Theorem implies that the derivative of the error at \(x_c\) is zero and the error is at its maximum at that point. In other words, to estimate the maximum error, we only need to find an upper bound of \(|R(x_c)|\).

We now perform a Taylor expansion of the error around \(x_c\). Using again the Cauchy form of the remainder, we find \[
R(x) = R(x_c) + xR'(x_c) + \frac{1}{2}R’^\prime(\xi_c)(x-\xi_c)(x-x_c).
\]
The second term on the right hand side is zero by construction, and we have \[
R(x) = R(x_c) + \frac{1}{2}R’^\prime(\xi_c)(x-\xi_c)(x-x_c).
\]
Let \(h\) again denote the distance between the two points, \(h=x_{i+1} – x_i\). We assume that \(x_c – x_i < h/2\) and use the equation above to calculate \(R(x_i)\) which we know is zero. If \(x_c\) was closer to \(x_{i+1}\) we would have to calculate \(R(x_{i+1})\) but otherwise the argument would remain the same. So, \[
R(x_i) = 0 = R(x_c) + \frac{1}{2}R’^\prime(\xi_c)(x_i-\xi_c)(x_i-x_c)
\]
from which we get \[
|R(x_c)| = \frac{1}{2}|R’^\prime(\xi_c)(x_i-\xi_c)(x_i-x_c)|.
\]
To get an upper estimate of the remainder that does not depend on \(x_c\) or \(\xi_c\) we can use the fact that both \(x_i-\xi_c \le h/2\) and \(x_i-x_c \le h/2\). We also know that \(|R(x)| \le |R(x_c)|\) over the interval from \(x_i\) to \(x_{i+1}\) and \(|R’^\prime(\xi_c)| = |f’^\prime(\xi_c)| \le |f’^\prime(x)|_{\mathrm{max}}\). Given all this, we end up with \[
|R(x)| \le \frac{h^2}{8}|f’^\prime(x)|_{\mathrm{max}}.
\]

The error of the linear interpolation scales with \(h^2\), in contrast to \(h\) for the piecewise constant interpolation. This means that increasing the number of samples gives us much more profit in terms of accuracy. Linear interpolation is often the method of choice because of its relative simplicity combined with reasonable accuracy. In a future post, I will be looking at higher-order interpolations. These higher-order schemes will scale even better with the number of samples but this improvement comes at a cost. We will see that the price to be paid is not only a higher computational expense but also the introduction of spurious oscillations that are not present in the original data.

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Frege’s Numbers

In a previous post, I started talking about natural numbers and how the Peano axioms define the relation between natural numbers. These axioms allow you to work with numbers and are good enough for most everyday uses. From a philosophical point of view, the Peano axioms have one big drawback. They only tell us how natural numbers behave but they don’t say anything about what natural numbers actually are. In the late 19th Century mathematicians started using set theory as the basis to define the axioms of arithmetic and other branches of mathematics. Two mathematicians, first Frege and later Bertrand Russell came up with a definition of natural numbers that gives us some insight into the nature of these elusive objects. In order to understand their definitions, I will first have to make the little excursion into set theory.

You may have encountered the basics of set theory already in primary school. Naïvely speaking sets are collections of things. Often the object in a set share some common property but this is not strictly necessary. You may have drawn Venn diagrams to depict sets, their unions and intersections. Something that is not taught in primary school is that you can define relations between sets that, in turn, define the so-called cardinality of a set.

Functions and Bijections

One of the central concepts is the mapping between two sets. For the following let’s assume we have two sets, \(\mathcal{A}\) and \(\mathcal{B}\). A function simply defines a rule that assigns an element of set \(\mathcal{B}\) to each element of set \(\mathcal{A}\). We call \(\mathcal{A}\) the domain of the function and \(\mathcal{B}\) the range of the function. If the function is named \(f\), then we write \[
f: \mathcal{A} \to \mathcal{B}
\]
to indicate what the domain and the range of the function are.

A function that maps vowels to uppercase letters

Example: For example, if \(\mathcal{A}\) is the set of uppercase and lowercase vowels, \[
\mathcal{A} = { A, E, I, O, U, a, e, i, o, u },
\]
and \(\mathcal{B}\) is the set of all uppercase letters in the alphabet, \[
\mathcal{B} = { A, B, C, D, \ldots, Z}.
\]

Now we can define a function that assigns the uppercase letter in \(\mathcal{B}\) to each vowel in \(\mathcal{A}\). So the mapping looks like shown in the figure.

You will notice two properties about this function. Firstly, not all elements from \(\mathcal{B}\) appear as a mapping of an element from \(\mathcal{A}\). We say that the uppercase consonants in \(\mathcal{B}\) are not in the image of \(\mathcal{A}\).

The second thing to note is that some elements in \(\mathcal{B}\) appear twice. For example, both the lowercase e and the uppercase E in \(\mathcal{A}\) map to the same uppercase E in \(\mathcal{B}\).

Definition of a Bijection

The example shows a function that is not a bijection. In order to be a bijection, a function must ensure that each element in the range is mapped to by exactly one element from the range. In other words for a function \[
f: \mathcal{A} \to \mathcal{B}
\]

  • every element in \(\mathcal{B}\) appears as a function value. No element is left out.
  • no element in \(\mathcal{B}\) appears as a function value more than once.

A bijection implies a one-to-one relationship between the elements in set \(\mathcal{A}\) and set \(\mathcal{B}\).

Equinumerosity and Cardinality

Intuitively, it is clear that you can only have a bijection between two sets if they have the same number of elements. After all, each element in \(\mathcal{A}\) is mapped onto exactly one element in \(\mathcal{B}\). This can be used to define a relation between any two sets.

Two sets are called equinumerous if there exists a bijection between the two sets. Equinumerous literally means “having the same number”. But we have to be careful here because we don’t yet know what the term “number” is supposed to mean. That is the reason why we define the term by using bijections and not referring to any “amount” or “number of elements”. Instead of saying that two sets are equinumerous, we can also say that they have the same cardinality.

Now comes the clever bit that Frege proposed. Let’s create a class of sets that all share the same cardinality. We can do that because equinumerosity is an equivalence relation but I won’t go into detail about what that means. We will call this cardinality class \(N\), so \[
N(\mathcal{A})
\]
is the class of all the sets that are equinumerous to \(\mathcal{A}\).

Intuitively we now have a class with all the sets that contain exactly one element, another class with all the sets that contain exactly two elements, and so forth. But we don’t know anything about numbers yet, so we also don’t really know what one and two are supposed to mean.

Constructing Natural Numbers

Now we have all the tools to construct the natural numbers \(\mathbb{N}\). Of course, we want our numbers to obey the Peano axioms, so we need two things. We need a zero element and we need a successor function \(S(n)\) that produces the next number from any given number.

The Zero Element

The zero-element is easily defined. We can construct the empty set, \[
\emptyset = \{\}.
\]
This is the set with no elements in it. Now the zero-element is simply the cardinality class of the empty set, \[
0 = N(\emptyset).
\]
This means that zero is a class of sets that all share the same cardinality as the empty set. You can show that this class consists of only one element, the empty set, by I won’t go into that here.

The Successor Function

Given that we have defined the zero element, \(0\), we can now define a set that contains zero as a single element, \[
\{0\}.
\]
Intuitively, this set has one element and we can thus define the natural number \(1\) as the cardinality class of this set, \[
1 = N(\{0\}).
\]
In general, given any natural number \(n\) we can define the successor \(S(n)\) by creating the cardinality class of the set that contains \(n\) together with all its predecessors, \[
n+1 = S(n) = N(\{0, 1, \ldots, n\}).
\]
You might think that this definition is somewhat circular. We are defining the successor function by using the concept of the predecessors. But this is not as problematic as it might seem at first sight. We know that the predecessor of \(1\) is \(0\) and each time we construct the next natural number, we can keep track of all the predecessors that we have constructed so far.

Conclusion

The zero and the successor function defined above are enough to define all the natural numbers \(\mathbb{N}\). I will not go into the proof that all the Peano axioms are satisfied by this construction. It is relatively straightforward and not very instructive in my opinion. If you want, you can try doing the proof as an exercise.

I personally find the Frege definition of natural numbers the most satisfying. It tells us that a number is not just some random symbol that doesn’t relate to the real world. A natural number is the class of all sets that share the same property. Each set in the class has the same cardinality and we can identify the cardinality with that number. It means that any set of objects in the real world can be thought of as an instance of a number. The number itself is the collection of sets and the concrete set is contained within it as an element. For example, if you see five apples on a table, you can think of them as a manifestation of the number \(5\).

This set of pins is an instance of the number 4 (four).

Another consequence of the definition of cardinality is that it gives us the ability to speak about infinities. A set might have an infinite number of elements. We already encountered \(\mathbb{N}\), the set of all natural numbers. Using the cardinality, we can compare infinite sets and create a hierarchy of infinities. I might talk about this more in a later post.

It would not be fair, however, if I didn’t mention some serious problems with the definition that I Frege came up with. The main problem arises because we are creating classes of sets without explicitly saying which elements we are allowing to be in those sets. This allows sets to contain arbitrary elements, including other sets. A set can even include itself as an element. This leads to the famous paradox by Russel which can be summarised as follows. Construct a set \(\mathcal{R}\) of all the sets that do not include themselves as an element. Then ask the question, does \(\mathcal{R}\) include itself? There are mathematical frameworks that attempt to save the essence of Frege’s definition of the natural numbers without running into these problems. In my personal opinion, they always lose some of the beauty and simplicity. But this is a necessary concession to make if you want to end up with a mathematical framework that doesn’t contain internal contradictions.

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What are Numbers? Or, Learning to Count!

We all use numbers every day and some of us feel more comfortable dealing with them than others. But have you ever asked yourself what numbers really are? For example, what is the number 4? Of course, you can describe the symbol “4”. But that is not really the number, is it? You can use Roman numerals IV, Urdu ۴, or Chinese and Japanese Kanji . Each one of these symbols represents the same number. And yet, somehow we would all probably agree that there is only one number 4.

The question about the nature of numbers is twofold. You can understand this question purely mathematical one and ask for a clear definition of a number and the set of numbers in a mathematical sense. This will be the topic of this article. You can also ask yourself what numbers are in a philosophical sense. Do numbers exist? If yes, in what way do they exist, and what are they? This may be the topic of a future article.

Now that we have settled what type of question we want to answer, we should start with the simplest type of numbers. These are the natural numbers 0, 1, 2, 3, 4, …
I decided to include the number zero even though it might seem a little abstract at first. After all, what does it mean to have zero of something? But that objection strays into the philosophical realm again and, as I said above, I want to focus on the mathematical aspect here.

When doing the research for this article, I was slightly surprised at the plethora of different definitions for the natural numbers. But given how fundamental this question is, it should be no wonder that many mathematicians have thought about the problem of defining numbers and have come up with different answers.

The Peano Axioms

Let’s start with an axiomatic definition of numbers called the Peano axioms. This is one of the earliest strict definitions of the natural numbers in the modern sense. It doesn’t really state what the natural numbers are but focuses on how they behave. We start with the set of natural numbers, which we call $\mathbb{N}$, and a successor function $S$.

Peano Axiom 1:

$0$ is a natural number or, more formally, $0 \in \mathbb{N}$

This axiom just tells us that there is a natural number zero. We could have chosen 1 as the starting point but this is arbitrary.

Peano Axiom 2:

Every natural number $x$ has a successor $y$.
In other words, given that $x \in \mathbb{N}$ then it follows that $y = S(x) \in \mathbb{N}$.

Intuitively, the successor function will always produce the next natural number.
Mathematicians call this property of the natural number being closed under the successor operation $S$. All this means is that the successor operation will never produce a result that is outside of the natural numbers.

Peano Axiom 3:

If we have two natural numbers $x$ and $y$, and we know that the successors of $x$ and $y$ are equal, then this is the same as saying that $x$ and $y$ are equal.
Again, written more formally we say, given $x \in \mathbb{N}$ and $y \in \mathbb{N}$ and $S(x) = S(y)$ then it follows that $x=y$.

This means that any natural number cannot be the successor of two different number. In other words, if you have two different numbers then they can’t have the same successor.

Peano Axiom 4:

$0$ is not the successor of any other natural number.
In mathematical notation, if $x \in \mathbb{N}$ then $S(x) \ne 0$

At first sight, we might think that these axioms are complete. We can start from zero and use the successor function $S$ to iterate through the natural numbers. This is intuitively shown in the image below.

Repeatedly applying the successor function, starting from 0.

Here
1 = S(0)
2 = S(S(0)) = S(1)
3 = S(S(S(0))) = S(2)
and so on.

But we haven’t guaranteed that this iteration will eventually be able to reach all natural numbers. Imagine that, in addition to the above, there is some special number $z$ that creates a closed loop when applying the successor function. And this loop is separate from the sequence that we get when starting from zero.

A closed loop of successors that isn’t connected to the zero element.

So we need another axiom that guarantees that all natural numbers are reachable from zero by repeatedly applying the successor function. This axiom is usually stated in the following way.

Peano Axiom 5: Axiom of Induction

Given any set $U \subseteq \mathbb{N}$ with $0 \in U$.
If $U$ is such that for every $n \in U$ the successor is also in $U$, i.e. $S(n) \in U$,
then $U = \mathbb{N}$

The idea behind this axiom is that $U$ can be chosen as the minimal set that contains no additional loops or numbers that are not reachable from zero. The fact that any set $U$ that contains zero and is closed under the successor function is identical to the natural numbers guarantees that all natural numbers are eventually reachable from zero. The axiom of induction is maybe more familiar in its alternative form.

Peano Axiom 5: Axiom of Induction, alternative form

Consider a mathematical statement that is true for zero.
If it can be proven that,
given the statement is true for a number $n$, then it is also true for $S(n)$,
then it follows that the statement is true for all natural numbers.

Here you can see that this axiom forms the basis of the familiar proof by induction.

Some Remarks

The Peano Axioms are helpful in defining the set of natural numbers $\mathbb{N}$ and arithmetic operations on them. But personally, I feel unsatisfied by this definition. The Peano Axioms tell us how natural numbers behave but they don’t really give any additional insight as to what numbers really are.

Take for example the number 2. We now know that 2 can be expressed as the successor of the successor of 0, i.e. $2 = S(S(0))$. We also know that this second successor must be a member of the set of natural numbers, but not much more. The problem here is that we never defined what the successor function should be.

Nonetheless, the Peano axioms can serve as a basis for more in-depth definitions of the natural numbers. These definitions can be considered models of the Peano Axioms in the sense that they define a zero element and some concrete successor function. The set of natural numbers can then be constructed from these and the Peano Axioms follow as consequences from these definitions.

In a future post, I will look at some set-theoretic definitions of the natural numbers. If you liked this post, please leave a comment and check back soon for more content.

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Playing around with Boost.GIL

One of the things that fascinate me about computers and algorithms is how they can turn a simple idea into artistic images. Even simple algorithms can produce unexpected outcomes. I used to experiment with algorithmic art in the past, but it never went much beyond the standard techniques that I read about on the internet. So, whatever I produced were mostly copies of other peoples ideas. I wanted to give myself a fresh start and try to come up with new and original imagery. This is going to be a process that will take some time. I am planning to first revisit the familiar techniques before moving on to more adventurous fields.

To start my journey along this path, I have decided to not rely on existing software but develop everything from scratch. So in some ways, I decided to re-invent the wheel for no other apparent reason than that I want to be in control. But I can’t really code everything from scratch. At some point, I will have to decide to rely on some solid base. Having chosen C++ as the language for my endeavour, I found Boost.GIL to be the perfect library for my purpose. It supplies an abstraction layer that allows me to work with colours and save my results in various images formats. But it doesn’t prescribe anything much else and I am free to create the images the way I want to. There are no routines to draw lines or shapes, no functions to blend layers, and no brushes. Just enough to assemble images pixel by pixel and write them to an output file.

The first component I decided to create is a colour map. This component simply maps a floating-point value to a colour. Using this I created the picture on the right. I simply calculated the distance from the centre and inserted this radius into my colour map. Choosing some interesting colour stops, a set of concentric rings is created. I admit it’s probably not a masterpiece but I gave it a name anyway. I called it “Floating Point” because of its appearance and in reference to the floating-point numerics used to create it.

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The SIR Model for the Spread of Infectious Diseases

In the current Coronavirus crisis, everybody is talking about flattening “the curve”. In the news, you will often see graphs of the total number of cases or the total number of deaths over time. So you may be forgiven to think that these are the curves that everybody is trying to flatten. In fact, what epidemiologists mean by the curve is the graph of the number of actively infected people over time. This curve is important because it determines the load that is placed on the healthcare system of a country. The current number of cases determines how many hospital beds, how many ventilators, and how much healthcare personnel are needed.

Mathematics and computer simulations play an important role in estimating how the disease will spread, how many people will be affected, and how much resources are needed. They also allow predicting the effects of different measures to control the spread. For example, the current lockdown in many countries around the world is reducing the number of people that an infected individual can pass the virus on to. It is important to know how effective this measure is. One of the big questions is when it is safe to relax the isolation of people and how much it would affect the spread if individual businesses re-open.

Before continuing, I have to add a disclaimer. I am interested in mathematics but I am not an expert epidemiologist. The models I am showing you here are very simple starting points for simulating the spread of diseases. They can give you some idea on how parameters like the infection rate and recovery rate influence the overall number of infected individuals. But they should not be used to draw any quantitative conclusions.

The SIR Model

To get a basic feel for the way infections spread through a population, epidemiologists have developed simple mathematical models. Probably the first model you will hear about in this context is the SIR model. The SIR model is a version of a compartmental model. This means that the total population is divided up into separate compartments. The quantity $S$ denotes the number of susceptible individuals. These are the people that are not infected and also don’t have any immunity to the disease. $I$ is the number of infected individuals and $R$ is the number of individuals that are not infectious but also can’t get the disease. Most scientists denote the $R$ to mean removed as it includes both people who have recovered and are immune but also those that have died. Due to the current sensitivity of the subject, many people prefer to call $R$ the recovered population.

Compartmental models define rates by which individuals change from one population to another. The SIR model has two rates, the rate of infection and the rate of recovery. The absolute rate of infection is proportional to the number of infected people. On average, each infected individual will pass the infection to a number of people in a given time interval. This number is usually called $\beta$. However, if an infected individual passes the virus to a recovered person, the infection will not spread. The probability of passing the infection on is given by $S/N$ where $N$ is the total population $N=S+I+R$. Putting this together, the absolute rate of infection is

$$\frac{\beta I S}{N}.$$

The rate of recovery is slightly more simple. Each infected individual will recover with some probability $\gamma$ in a given time interval. The absolute rate of recovery is then expressed as

$$\gamma I.$$
The infection rate reduces the number of susceptible individuals $S$ and increases the number of infected individuals $I$. The recovery rate reduces the number of infected individuals $I$ and increases the number of recovered individuals $R$. The complete set of rate equations is then

$$\begin{eqnarray}
\frac{dS}{dt} &=& – \frac{\beta I S}{N}, \\
\frac{dI}{dt} &=& \frac{\beta I S}{N} – \gamma I, \\
\frac{dR}{dt} &=& \gamma I.
\end{eqnarray}$$

The ratio of the coefficients $\beta$ and $\gamma$ is known as the basic reproduction ratio.

$$R_0 = \frac{\beta}{\gamma}$$.

The $R_0$ is important because it determines whether the infection will spread exponentially or eventually die out.

I have implemented a little JavaScript app that integrates the SIR equations and shows the development of the populations over time. Feel free to play around with the sliders and explore how the parameters influence the spread.

Please enable JavaScript

I encourage you to play around with the parameters to see how the model behaves. For an infection rate of 1 and a recovery rate of 0.5, the populations stabilise when about 80% of the population has been infected and has recovered. The maximum of the infectious population, the $I$ curve, reaches about 16%. If you reduce the infection rate, the $I$ curve flattens, prolonging the time over which the disease is spreading but reducing the maximum number of infected individuals at any one time.

The SEIR Model

One of the major assumptions in the SIR model is that an infected individual can immediately spread the infection. A refinement of the model is the addition of a population, $E$, of exposed individuals. These are people that are infected but are not yet infectious. The SEIR model introduces another rate, $a$, at which exposed individuals turn infectious. The quantity $a$ can be understood as the inverse of the average incubation period. The absolute rate at which exposed individuals become infectious is

$$a E.$$

The complete set of equations of the SEIR model are then as follows.

$$\begin{eqnarray}
\frac{dS}{dt} &=& – \frac{\beta I S}{N}, \\
\frac{dE}{dt} &=& \frac{\beta I S}{N} – a E, \\
\frac{dI}{dt} &=& a E – \gamma I, \\
\frac{dR}{dt} &=& \gamma I.
\end{eqnarray}$$

The SEIR model is also implemented in the app above. Simply pick SEIR Model from the dropdown menu and start exploring.

The SEIR Model with Delay

The SEIR model above assumes that an individual, once exposed, can immediately turn infectious. The constant rate $a$ implies that the probability of changing from the exposed state to the infectious state is the same on day one of being exposed as it is on day ten. This might not be realistic because diseases typically have some incubation period. Only after some number of days after being exposed will an individual become infectious. One can model this kind of behaviour with a time delay. Let’s say that after a given incubation period $\tau$, every exposed individual will turn infectious. The absolute rate at which exposed individuals become infectious is then given by

$$\frac{\beta I(t-\tau) S(t-\tau)}{N}.$$

Here the $S(t-\tau)$ means taking the value of the susceptible individuals not at the current time, but at a time in the past with a delay of $\tau$. The complete set of equations of the SEIR model with delay are then as follows.

$$\begin{eqnarray}
\frac{dS}{dt} &=& – \frac{\beta I(t) S(t)}{N}, \\
\frac{dE}{dt} &=& \frac{\beta I(t) S(t)}{N} – \frac{\beta I(t-\tau) S(t-\tau)}{N}, \\
\frac{dI}{dt} &=& \frac{\beta I(t-\tau) S(t-\tau)}{N} – \gamma I(t), \\
\frac{dR}{dt} &=& \gamma I(t).
\end{eqnarray}$$

I have written the time dependence explicitly for all quantities on the right-hand side to make it clear how the time delay should be applied.

You can choose this model in the app above by selecting SEIR Model with Delay from the dropdown menu.

Some Conclusions

The SEIR model and the SEIR model with delay both introduce a population of exposed individuals that are not yet infectious. This draws out the spread of the disease over a longer time. It also slightly reduces the maximum of the infectious population curve $I$. Introducing a time delay doesn’t change the curves too much. But for long incubation periods, the curve of infectious individuals can have multiple maxima. So at some time, it may look like the disease has stopped spreading while in reality, a next wave is just about to start. The two versions of the SEIR model are two extremes and the truth lies somewhere in between these two.

I have to stress again that I am not an epidemiology expert and that the models presented here are very simple models. For any meaningful prediction of the spread of a real disease, much more complex models are needed. These models must include real data about the number of contacts that different parts of the population have between each other.

The code for the application above is available on
GitHub

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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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