Computational Physics Basics: Polynomial Interpolation

Posted 19th April 2023 by Holger Schmitz

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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Starting GPU programming with Kokkos

Posted 9th March 2023 by Holger Schmitz

The high performance computing landscape has changed quite a bit over the last years. For quite a few decades, the major architecture for HPC systems was based on classical CPU processors distributed over many nodes. The nodes are connected via a fast network architecture. These compute clusters have evolved over time. A typical simulation code that makes use of this architecture will consist of more or less serial code that runs on each core and communicates with the other cores via the message passing library MPI. This has the drawback that the serial instances of the simulation code running on a shared memory node still have to exchange data through this mechanism, which takes time and reduces efficiency. On many current systems, each node can have on the order of 10s of compute cores accessing the same shared memory. To make better use of these cores, multithreading approaches, such as OpenMP, have been used increasingly over the past decade or so. However, there are not many large scale simulation codes that make use of OpenMP and MPI at the same time to reach maximum performance.

More recently, there has been new style of architecture that has become a serious player. GPU processors that used to play more of an outsider role in serious high performance computing have taken centre stage. Within the DoE’s Exascale project in the US, new supercomputers are being built that are breaking the exaflop barrier. These machines rely on a combination of traditional CPU cluster with GPU accelerators.

For developers of high performance simulation codes this means that much of their code has to be re-written to make use of the new mixed architectures. I have been developing the Schnek library for a number of years now and I am using it form many of my simulation codes. To keep the library relevant for the future, I am currently in the process of adding mixed architecture capabilities to Schnek. Fortunately, there already exists the Kokkos library which provides an abstraction layer over the different machine architectures. It also provides multi-dimensional arrays that can be stored on the host memory or on the accelerator’s device memory. It also provides iteration mechanisms to process these arrays on the CPU or GPU.

In this post, I am describing my first experiences with Kokkos and how it can be used to accelerate calculations. I will be starting from scratch which means that I will begin by installing Kokkos. The best way to do this nowadays is to use a packet manager called Spack. Spack is designed as a packet manager for supercompters but will work on any regular Linux machine. To start, I navigate into a folder where I want to install spack, let’s call it /my/installation/folder. I then simply clone Spack’s git repository.

git clone -c feature.manyFiles=true https://github.com/spack/spack.git

I am using the bash shell, so now I can initialise all the environment variables by running

source /my/installation/folder/spack/share/spack/setup-env.sh

This command should be run every time you want to start developing code using the packages supplied by Spack. I recommend not putting this line into your .bashrc because some of the Spack packages might interfere with your regular system operation.

The Kokkos package has many configuration options. You can now look at them using the command

spack info kokkos

I have an NVidia Quadro P1000 graphics card on my local development machine. For NVidia accelerators, I will need to install Kokkos with Cuda support. In order to install the correct Cuda version, I first check on https://developer.nvidia.com/cuda-gpus for the compute capability. For my hardware, I find that I will need version 6.1. The following command will install Kokkos and all of its dependencies.

spack install kokkos +cuda +wrapper cuda_arch=61

The wrapper option is required because I am compiling everything with the gcc compiler. Note that this can take quite some time because many of the packages may be build from source code during the installation. This allows Spack packages to be highly specific to your system.

spack load kokkos

Now, I can start creating my first Kokkos example. I create a file called testkokkos.cpp and add some imports to the top.

#include <Kokkos_Core.hpp>
#include <chrono>

The Kokkos_Core.hpp import is needed to use Kokkos and I also included chrono from the standard library to allow timing the code. Kokkos introduces execution spaces and memory spaces. Data can live on the host machine or the accelerator device. Similarly, execution of the code can be carried out on the serial CPU or the GPU or some other specialised processor. I want my code to be compiled for two different settings so that I can compare GPU performance against the CPU. I define two types Execution and Memory for the execution and memory spaces. These types will depend on an external macro that will be passed in by the build system.

#ifdef TEST_USE_GPU
    typedef Kokkos::Cuda Execution;
    typedef Kokkos::CudaSpace Memory;
#else
    typedef Kokkos::Serial Execution;
    typedef Kokkos::HostSpace Memory;
#endif

Kokkos manages data in View objects which represent multi-dimensional arrays. View has some template arguments. The first argument specifies the dimensionality and the datatype stored in the array. Further template arguments can be specified to specify the memory space and other compile-time configurations. For example Kokkos::View<double**, Kokkos::HostSpace> defines a two dimensional array of double precision values on the host memory. To iterate over a view, one needs to define function objects that will be passed to a function that Kokkos calls “parallel dispatch”. The following code defines three such structs that can be used to instantiate function objects.

#define HALF_SIZE 500

struct Set {
    Kokkos::View<double**, Memory> in;

    KOKKOS_INLINE_FUNCTION void operator()(int i, int j) const {
        in(i, j) = i==HALF_SIZE && j==HALF_SIZE ? 1.0 : 0.0;
    }
};

struct Diffuse {
    Kokkos::View<double**, Memory> in;
    Kokkos::View<double**, Memory> out;

    KOKKOS_INLINE_FUNCTION void operator()(int i, int j) const {
        out(i, j) = in(i, j) + 0.1*(in(i-1, j) + in(i+1, j) + in(i, j-1) + in(i, j+1)  - 4.0*in(i, j));
    }
};

The Set struct will initialise an array to 0.0 everywhere except for one position where it will be set to 1.0. This results in a single spike in the centre of the domain. Diffuse applies a diffusion operator to the in array and stored the result in the out array. The calculations can’t be carried out in-place because the order in which the function objects are called may be arbitrary. This means that, after the diffusion operator has been applied, the values should be copied back from the out array to the in array.

Now that these function objects are defined, I can start writing the actual calculation.

void performCalculation() {
    const int N = 2*HALF_SIZE + 1;
    const int iter = 100;

    Kokkos::View<double**, Memory> in("in", N, N);
    Kokkos::View<double**, Memory> out("out", N, N);

    Set set{in};
    Diffuse diffuse{in, out};

The first two line in the function define some constants. N is the size of the grids and iter sets the number of times that the diffusion operator will be applied. The Kokkos::View objects in and out store the 2-dimensional grids. The first template argument double ** specifies that the arrays are 2-dimensional and store double values. The Memory template argument was defined above and can either be Kokkos::CudaSpace or Kokkos::HostSpace. The last two lines in the code segment above initialise my two function objects of type Set and Diffuse.

I want to iterate over the inner domain, excluding the grid points at the edge of the arrays. This is necessary because the diffusion operator accesses the grid cells next to the position that is iterated over. The iteration policy uses the multidmensional range policy from Kokkos.

    auto policy = Kokkos::MDRangePolicy<Execution, Kokkos::Rank<2>>({1, 1}, {N-1, N-1});

The Execution template argument was defined above and can either be Kokkos::Cuda or Kokkos::Serial. The main calculation now looks like this.

    Kokkos::parallel_for("Set", policy, set);
    Kokkos::fence();

    auto startTime = std::chrono::high_resolution_clock::now();

    for (int i=0; i<iter; ++i)
    {
        Kokkos::parallel_for("Diffuse", policy, diffuse);
        Kokkos::fence();
        Kokkos::deep_copy(in,  out);
    }

    auto endTime = std::chrono::high_resolution_clock::now();
    auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(endTime - startTime).count();

    std::cerr << "Wall clock: " <<  milliseconds << " ms" << std::endl;
}

The function Kokkos::parallel_for applies the function object for each element given by the iteration policy. Depending on the execution space, the calculations are performed on the CPU or the GPU. To set up the calculation the set operator is applied. Inside the main loop, the diffusion operator is applied, followed by a Kokkos::deep_copy which copies the out array back to the in array. Notice, that I surrounded the loop by calls to STL’s high_resolution_clock::now(). This will allow me to print out the wall-clock time used by the calculation and give me some indication of the performance of the code.

The main function now looks like this.

int main(int argc, char **argv) {
    Kokkos::initialize(argc, argv);

    performCalculation();    

    Kokkos::finalize_all();
    return 0;
}

It is important to initialise Kokkos before any calls to its routines are called, and also to finalise it before the program is exited.

To compile the code, I use CMake. Kokkos provides the necessary CMake configuration files and Spack sets all the paths so that Kokkos is easily found. My CMakeLists.txt file looks like this.

cmake_minimum_required(VERSION 3.10)

project(testkokkos LANGUAGES CXX)

set(CMAKE_CXX_STANDARD 17)
set(CMAKE_CXX_STANDARD_REQUIRED ON)
set(CMAKE_CXX_EXTENSIONS OFF)

set(CMAKE_ARCHIVE_OUTPUT_DIRECTORY ${CMAKE_BINARY_DIR}/lib)
set(CMAKE_LIBRARY_OUTPUT_DIRECTORY ${CMAKE_BINARY_DIR}/lib)
set(CMAKE_RUNTIME_OUTPUT_DIRECTORY ${CMAKE_BINARY_DIR}/bin)

find_package(Kokkos REQUIRED PATHS ${KOKKOS_DIR})
# find_package(CUDA REQUIRED)

add_executable(testkokkosCPU testkokkos.cpp)
target_compile_definitions(testkokkosCPU PRIVATE KOKKOS_DEPENDENCE)
target_link_libraries(testkokkosCPU PRIVATE Kokkos::kokkoscore)

add_executable(testkokkosGPU testkokkos.cpp)
target_compile_definitions(testkokkosGPU PRIVATE TEST_USE_GPU KOKKOS_DEPENDENCE)
target_link_libraries(testkokkosGPU PRIVATE Kokkos::kokkoscore)

Running cmake . followed by make will build two targets, testkokkosCPU and testkokkosGPU. In the compilation of testkokkosGPU the TEST_USE_GPU macro has been set so that it will make use of the Cuda memory and execution spaces.

Now its time to compare. Running testkokkosCPU writes out

Wall clock: 9055 ms

Running it on the GPU with testkokkosGPU gives me

Wall clock: 57 ms

That’s right! On my NVidia Quadro P1000, the GPU accelerated code outperforms the serial version by a factor of more than 150.

Using Kokkos is a relatively easy and very flexible way to make use of GPUs and other accelerator architectures. As a developer of simulation codes, the use of function operators may seem a bit strange at first. From a computer science perspective, on the other hand, these operators feel like a very natural way of approaching the problem.

My next task now is to integrate the Kokkos approach with Schnek’s multidimensional grids and fields. Fortunately, Schnek provides template arguments that allow fine control over the memory model and this should allow replacing Schnek’s default memory model with Kokkos views.

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Computational Physics Basics: Piecewise and Linear Interpolation

Posted 24th February 2022 by Holger Schmitz

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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Computational Physics: Truncation and Rounding Errors

Posted 15th October 2021 by Holger Schmitz

In a previous post, I talked about accuracy and precision in numerical calculations. Ideally one would like to perform calculations that are perfect in these two aspects. However, this is almost never possible in practical situations. The reduction of accuracy or precision is due to two numerical errors. These errors can be classified into two main groups, round-off errors and truncation errors.

Rounding Error

Round-off errors occur due to the limits of numerical precision at which numbers are stored in the computer. As I discussed here a 32-bit floating-point number for example can only store up to 7 or 8 decimal digits. Not just the final result but every intermediate result of a calculation will be rounded to this precision. In some cases, this can result in a much lower precision of the final result. One instance where round-off errors can become a problem happens when the result of a calculation is given by the difference of two large numbers.

Truncation Error

Truncation errors occur because of approximations the numerical scheme makes with respect to the underlying model. The name truncation error stems from the fact that in most schemes the underlying model is first expressed as an infinite series which is then truncated allowing it to be calculated on a computer.

Example: Approximating Pi

Let’s start with a simple task. Use a series to approximate the value of \(\pi\).

Naive summation

One of the traditional ways of calculating \(\pi\) is by using the \(\arctan\) function together with the identity \[
\arctan(1) = \frac{\pi}{4}.
\] One can expand \(\arctan\) into its Taylor series, \[
\arctan(x)
= x – \frac{x^3}{3} +\frac{x^5}{5} – \frac{x^7}{7} + \ldots
= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}.
\] The terms of the series become increasingly smaller and you could try to add up all the terms up to some maximum \(N\) in the hope that the remaining infinite sum is small and can be neglected. Inserting \(x=1\) into the sum will give you an approximation for \(\pi\), \[
\pi \approx 4\sum_{n=0}^N \frac{(-1)^n }{2n+1}.
\]

Here are e implementations for this approximation in C++, Python and JavaScript.

C++

double pi_summation_slow(int N) {
  double  sum = 0.0;
  int  sign = 1;

  for (int  i=0; i<N; ++i) {
    sum += sign/(2*i + 1.0);
    sign = -sign;
  }

  return 4*sum;
}

Python

def  pi_summation_slow(N):
  sum  =  0.0
  sign  =  1

  for  i  in  range(0,N):
    sum   =  sum  +  sign/(2*i  +  1.0)
    sign  =  -sign
    return  4*sum

JavaScript

function pi_summation_slow(N) {
  let sum = 0.0;
  let sign = 1;

  for (let i=0; i<N; ++i) {
    sum += sign/(2*i + 1.0);
    sign = -sign;
  }

  return 4*sum;
}

Let’s call this function with \(N=10\). All the results I am showing here are calculated using a Python implementation. We get a result of around 3.0418. The relative error is 0.0318 and is, of course, unacceptable. This error falls under the category of truncation errors because it is caused by not summing up enough terms of the Taylor series. Calling the function with \(N=1000\) gives us a result of 3.14059 with a relative error of \(3.183\times 10^{-4}\). The error has improved but is still far off from the possible \(10^{-14}\) to \(10^{-15}\) achievable in double-precision arithmetic. The figure below shows how the relative error decreases with the number of iterations.

Relative error of the simple approximation of \( \pi \) depending on the number of iterations

From this curve, one h long m wl hat the error decreases with \(1/N\). If one extrapolates the curve, one finds that it would take \(10^{14}\) iterations to reach an error below \(10^{-14}\). Even if this was computationally feasible, the round-off errors of such a long sum would eventually prevent the error from being lowered to this limit.

Improvements using Machin’s formula

The technique of calculating \(\pi\) can be improved in two ways. Firstly, instead of using the Taylor series, you can use Euler’s series for the \(\arctan\) function.

\[
\arctan(x) = \sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n + 1)!} \frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.
\]

This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for \(\pi\), first discovered by John Machin in 1706, \[
\frac{\pi}{4} = 4 \arctan \frac{1}{5} – \arctan \frac{1}{239}
\] Here are the implementations for this formula.

C++

double pi_summation_fast(int order) {
  using boost::math::factorial;
  double sum = 0.0;

  for (unsigned int n=0; n<order; ++n) {
    double f = factorial<double>(n);
    double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1);
    double A = pow(25./26., n+1)/pow(5., 2*n+1);
    double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1);
    sum += common*( 4*A - B );
  }

  return 4*sum;
}

Python

def pi_summation_fast(N):
  sum = 0.0

  for n in range(0,N):
    f = factorial(n)
    common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1)
    A = math.pow(25/26, n+1)/math.pow(5, 2*n+1)
    B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1)
    sum = sum + common*( 4*A - B )

  return 4*sum;

JavaScript

function pi_summation_fast(N) {
  let sum = 0.0;

  for (let n=0; n<N; ++n) {
    const f = factorial(n);
    const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1);
    const A = pow(25/26, n+1)/pow(5, 2*n+1);
    const B = pow(239*239 / (239*239 + 1), n+1)/pow(239, 2*n+1);
    sum += common*( 4*A - B );
  }

  return 4*sum;
}

The table below shows the computed values for \(\pi\) together with the relative error. You can see that each iteration reduces the error by more than an order of magnitude and only a few iterations are necessary to achieve machine precision accuracy.

Example: Calculating sin(x)

Calculate the value of \(\sin(x)\) using it’s Taylor series around x=0.

The Taylor series for \(\sin(x)\) is \[
\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}.
\] this series is much more well-behaved than the Taylor series for \(\arctan\) we saw above. Because of the factorial in the denominator, the individual terms of this series will converge reasonably quickly. Here are some naive implementations of this function where the infinite sum has been replaced by a sum from zero to \(N\).

C++

double taylor_sin(double x, int order)
{
  using boost::math::factorial;
  double sum = 0.0;
  int sign = 1;

  for (unsigned int n=0; n<order; ++n)
  {
    sum += sign*pow(x, 2*n + 1)/factorial<double>(2*n +1);
    sign = -sign;
  }

  return sum;
}

Python

def taylor_sin(x, N):
  sum = 0.0
  sign = 1

  for n in range(0,N):
    sum = sum + sign*math.pow(x, 2*n + 1)/factorial(2*n + 1)
    sign = -sign

  return sum

JavaScript

function taylor_sin(x, N) {
  let sum = 0.0;
  let sign = 1;

  for (let n=0; n<N; n++) {
    sum += sign*pow(x, 2*n + 1)/factorial(2*n +1);
    sign = -sign;
  }

  return sum;
}

A good test for this function is the evaluation of \(\sin(x)\) at values \(x = k\pi\), where \(k\) is an integer. We know that \(\sin(k\pi) = 0\) and the return value from the numeric function can directly be used as the absolute error of the computation. The figure below shows results for some values of \(k\) plotted against \(N\).

For small values of \(k\), this series converges relatively quickly. But for larger \(k\) you can see that more and more terms are needed. The error even grows first before being reduced. Just like the example above, the truncation error requires large values of \(N\) to reach a good accuracy of the result. In practice, you would not calculate the \(\sin\) function this way. Instead you would make use of known properties, such as \(\sin(2k\pi + x) = \sin(x)\) for integer \(k\), to transform the argument into a range where fast convergence is guaranteed.

However, I would like to continue my analysis of this function because it shows two more interesting pitfalls when performing long sums. First, you will notice that the curves in the figure above show dashed lines for \(N>85\). This is because the implementation I showed above will actually fail with a range error. The pow function and the factorial both start producing numbers that exceed the valid range of double floating-point numbers. The quotient of the two, on the other hand, remains well-behaved. It is, therefore, better to write the Taylor series using a recursive definition of the terms.

\[
\sin x = \sum_{n=0}^\infty a_n(x),
\] with \[
a_0 = x
\] and \[
a_{n} = -\frac{x^2}{2n(2n+1)}a_{n-1}
\]

The implementations are given again below.

C++

double taylor_sin_opt(double x, int order)
{
  double sum = x;
  double an = x;

  for (unsigned int n=1; n<order; ++n)
  {
    an = -x*x*an/(2*n*(2*n+1));
    sum += an;
  }

  return sum;
}

Python

def taylor_sin_opt(x, N):
  sum = x
  an = x

  for n in range(1,N):
    an = -x*x*an/(2*n*(2*n+1))
    sum = sum + an

  return sum

JavaScript

function taylor_sin_opt(x, N) {
  let sum = x;
  let an = x;

  for (let n=1; n<N; n++) {
    an = -x*x*an/(2*n*(2*n+1));
    sum += an;
  }

  return sum;
}

The other takeaway from the graphs of the errors is that they don’t always converge to machine accuracy. The reason for this originates from fact that the initial terms of the sum can be quite large but with opposite signs. They should cancel each other out exactly, but they don’t because of numerical round-off errors.

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Computational Physics Basics: Accuracy and Precision

Posted 24th August 2021 by Holger Schmitz

Problems in physics almost always require us to solve mathematical equations with real-valued solutions, and more often than not we want to find functional dependencies of some quantity of a real-valued domain. Numerical solutions to these problems will only ever be approximations to the exact solutions. When a numerical outcome of the calculation is obtained it is important to be able to quantify to what extent it represents the answer that was sought. Two measures of quality are often used to describe numerical solutions: accuracy and precision. Accuracy tells us how will a result agrees with the true value and precision tells us how reproducible the result is. In the standard use of these terms, accuracy and precision are independent of each other.

Accuracy

Accuracy refers to the degree to which the outcome of a calculation or measurement agrees with the true value. The technical definition of accuracy can be a little confusing because it is somewhat different from the everyday use of the word. Consider a measurement that can be carried out many times. A high accuracy implies that, on average, the measured value will be close to the true value. It does not mean that each individual measurement is near the true value. There can be a lot of spread in the measurements. But if we only perform the measurement often enough, we can obtain a reliable outcome.

Precision

Precision refers to the degree to which multiple measurements agree with each other. The term precision in this sense is orthogonal to the notion of accuracy. When carrying out a measurement many times high precision implies that the outcomes will have a small spread. The measurements will be reliable in the sense that they are similar. But they don’t necessarily have to reflect the true value of whatever is being measured.

Accuracy vs Precision

Data with high accuracy but low precision. The green line represents the true value.

To fully grasp the concept of accuracy vs precision it is helpful to look at these two plots. The crosses represent measurements whereas the line represents the true value. In the plot above, the measurements are spread out but they all lie around the true value. These measurements can be said to have low precision but high accuracy. In the plot below, all measurements agree with each other, but they do not represent the true value. In this case, we have high precision but low accuracy.

Data with high precision but low accuracy. The green line represents the true value.

A moral can be gained from this: just because you always get the same answer doesn’t mean the answer is correct.

When thinking about numerical methods you might object that calculations are deterministic. Therefore the outcome of repeating a calculation will always be the same. But there is a large class of algorithms that are not quite so deterministic. They might depend on an initial guess or even explicitly on some sequence of pseudo-random numbers. In these cases, repeating the calculation with a different guess or random sequence will lead to a different result.

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