**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

Date: **Aug 25, 2017**

Copyright 1999-2017, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

**and discussed at the lab sessions the first two weeks.**

GIT for version control, discussed at the lab this week (and next week as well)

ipython notebook, mentioned this week

QTcreator for editing and mastering computational projects

Armadillo as a useful numerical library for C++, highly recommended

Unit tests

Before writing a single line, have the algorithm clarified and understood. It is crucial to have a logical structure of e.g., the flow and organization of data before one starts writing.

Always try to choose the simplest algorithm. Computational speed can be improved upon later.

Try to write a as clear program as possible. Such programs are easier to debug, and although it may take more time, in the long run it may save you time. If you collaborate with other people, it reduces spending time on debuging and trying to understand what the codes do. A clear program will also allow you to remember better what the program really does!

The planning of the program should be from top down to bottom, trying to keep the flow as linear as possible. Avoid jumping back and forth in the program. First you need to arrange the major tasks to be achieved. Then try to break the major tasks into subtasks. These can be represented by functions or subprograms. They should accomplish limited tasks and as far as possible be independent of each other. That will allow you to use them in other programs as well.

Try always to find some cases where an analytical solution exists or where simple test cases can be applied. If possible, devise different algorithms for solving the same problem. If you get the same answers, you may have coded things correctly or made the same error twice or more.

**Compiling and linking, without QTcreator.**

In order to obtain an executable file for a C++ program, the following instructions under Linux/Unix can be used

```
c++ -c -Wall myprogram.cpp
c++ -o myprogram myprogram.o
```

where the compiler is called through the command c++/g++. The compiler
option -Wall means that a warning is issued in case of non-standard
language. The executable file is in this case `myprogram`

. The option
`-c`

is for compilation only, where the program is translated into machine code,
while the `-o`

option links the produced object file `myprogram.o`

and produces the executable `myprogram`

.

For Fortran2008 we use the Intel compiler, replace `c++`

with `ifort`

.
Also, to speed up the code use compile options like

` c++ -O3 -c -Wall myprogram.cpp`

```
# Comment lines
# General makefile for c - choose PROG = name of given program
# Here we define compiler option, libraries and the target
CC= g++ -Wall
PROG= myprogram
# this is the math library in C, not necessary for C++
LIB = -lm
# Here we make the executable file
${PROG} : ${PROG}.o
${CC} ${PROG}.o ${LIB} -o ${PROG}
# whereas here we create the object file
${PROG}.o : ${PROG}.c
${CC} -c ${PROG}.c
```

If you name your file for `makefile`

, simply type the command
`make`

and Linux/Unix executes all of the statements in the above
makefile. Note that C++ files have the extension `.cpp`

.

**The C encounter.**

Here we present first the C version.

```
/* comments in C begin like this and end with */
#include <stdlib.h> /* atof function */
#include <math.h> /* sine function */
#include <stdio.h> /* printf function */
int main (int argc, char* argv[])
{
double r, s; /* declare variables */
r = atof(argv[1]); /* convert the text argv[1] to double */
s = sin(r);
printf("Hello, World! sin(%g)=%g\n", r, s);
return 0; /* success execution of the program */
```

` #include <stdlib.h> /* atof function */`

` int main (int argc, char* argv[])`

` int main (int argc, char* argv[])`

The integer `argc`

is the no of command-line arguments, set to
one in our case, while
`argv`

is a vector of strings containing the command-line arguments
with `argv[0]`

containing the name of the program
and `argv[1]`

, `argv[2]`

, ... are the command-line args, i.e., the number of
lines of input to the program.
Here we define floating points, see also below,
through the keywords `float`

for single precision real numbers and
`double`

for double precision. The function
`atof`

transforms a text (`argv[1]`

) to a float.
The sine function is declared in math.h, a library which
is not automatically included and needs to be linked when computing
an executable file.

With the command `printf`

we obtain a formatted printout.
The `printf`

syntax is used for formatting output
in many C-inspired languages (Perl, Python, Awk, partly C++).

**Now in C++.**

Here we present the C++ version using namespace.

```
// A comment line begins like this in C++ programs
// Standard ANSI-C++ include files
#include <iostream> // input and output
#include <cmath> // math functions
using namespace std;
int main (int argc, char* argv[])
{
// convert the text argv[1] to double using atof:
double r = atof(argv[1]); // convert the text argv[1] to double
double s = sin(r);
cout << "Hello, World! sin(" << r << ") =" << s << endl;
return 0; // success execution of the program
}
```

**Without namespace.**

Namespaces provide a method for preventing name conflicts in large projects. Symbols declared inside a namespace block are placed in a named scope that prevents them from being mistaken for identically-named symbols in other scopes. Multiple namespace blocks with the same name are allowed. All declarations within those blocks are declared in the named scope.

Here we present the C++ version without using namespace.

```
// A comment line begins like this in C++ programs
// Standard ANSI-C++ include files
#include <iostream> // input and output
#include <cmath> // math functions
using namespace std;
int main (int argc, char* argv[])
{
// convert the text argv[1] to double using atof:
double r = atof(argv[1]); // convert the text argv[1] to double
double s = sin(r);
// Note std::cout and std::endl
std::cout << "Hello, World! sin(" << r << ") =" << s << std::endl;
return 0; // success execution of the program
}
```

**Dissection I.**

We have replaced the call to `printf`

with the standard C++ function
`cout`

. The header file `<iostream.h>`

is then needed.
In addition, we don't need to
declare variables like `r`

and `s`

at the beginning of the program.
I personally prefer
however to declare all variables at the beginning of a function, as this
gives *me* a feeling of greater readability.

**C/C++ program.**

A C/C++ program begins with include statements of header files (libraries,intrinsic functions etc)

Functions which are used are normally defined at top (details next week)

The main program is set up as an integer, it returns 0 (everything correct) or 1 (something went wrong)

Standard

`if`

,`while`

and`for`

statements as in Java, Fortran, Python...Integers have a very limited range.

**Arrays.**

A C/C++ array begins by indexing at 0!

Array allocations are done by size, not by the final index value.If you allocate an array with 10 elements, you should index them from $0,1,\dots, 9$.

Initialize always an array before a computation.

**Integer and Real Numbers.**

Overflow

Underflow

Roundoff errors

Loss of precision

**C++ and Fortran declarations.**

type in C/C++ and Fortran2008 | bits | range |
---|---|---|

`int/INTEGER` (2) | 16 | -32768 to 32767 |

`unsigned int` | 16 | 0 to 65535 |

`signed int` | 16 | -32768 to 32767 |

`short int` | 16 | -32768 to 32767 |

`unsigned short int` | 16 | 0 to 65535 |

`signed short int` | 16 | $-32768$ to 32767 |

`int/long int/INTEGER` (4) | 32 | -2147483648 to 2147483647 |

`signed long int` | 32 | -2147483648 to 2147483647 |

`float/REAL(4)` | 32 | $3.4\times 10^{-44}$ to $3.4\times 10^{+38}$ |

`double/REAL(8)` | 64 | $1.7\times 10^{-322}$ to $1.7\times 10^{+308}$ |

`long double` | 64 | $1.7\times 10^{-322}$ to $1.7\times 10^{+308}$ |

**How to do it.**

$$
a_n2^n+a_{n-1}2^{n-1} +a_{n-2}2^{n-2} +\dots +a_{0}2^{0}.
$$

In binary notation we have thus $(417)_{10} =(110110001)_2$ since we have

$$
\begin{align*}
(110100001)_2
&=1\times2^8+1\times 2^{7}+0\times 2^{6}+1\times 2^{5}+0\times 2^{4}+0\times 2^{3}\\
&+0\times 2^{2}+0\times 2^{2}+0\times 2^{1}+1\times 2^{0}.
\end{align*}
$$

To see this, we have performed the following divisions by 2

417/2=208 | remainder 1 | coefficient of $2^{0}$ is 1 |

208/2=104 | remainder 0 | coefficient of $2^{1}$ is 0 |

104/2=52 | remainder 0 | coefficient of $2^{2}$ is 0 |

52/2=26 | remainder 0 | coefficient of $2^{3}$ is 0 |

26/2=13 | remainder 1 | coefficient of $2^{4}$ is 0 |

13/2= 6 | remainder 1 | coefficient of $2^{5}$ is 1 |

6/2= 3 | remainder 0 | coefficient of $2^{6}$ is 0 |

3/2= 1 | remainder 1 | coefficient of $2^{7}$ is 1 |

1/2= 0 | remainder 1 | coefficient of $2^{8}$ is 1 |

**Integer numbers.**

```
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstdlib>
using namespace std;
int main (int argc, char* argv[])
{
int i;
int terms[32]; // storage of a0, a1, etc, up to 32 bits
int save;
int number = atoi(argv[1]);
save = number;
// initialise the term a0, a1 etc
for (i=0; i < 32 ; i++){ terms[i] = 0;}
for (i=0; i < 32 ; i++){
terms[i] = number%2;
number /= 2;
}
// write out results
cout << "Number of bytes used= " << sizeof(number) << endl;
for (i=0; i < 32 ; i++){
cout << " Term nr: " << i << "Value= " << terms[i];
cout << endl;
}
return 0;
}
```

**Integer numbers, Fortran.**

```
PROGRAM binary_integer
IMPLICIT NONE
INTEGER i, number, terms(0:31) ! storage of a0, a1, etc, up to 32 bits
WRITE(*,*) 'Give a number to transform to binary notation'
READ(*,*) number
! Initialise the terms a0, a1 etc
terms = 0
! Fortran takes only integer loop variables
DO i=0, 31
terms(i) = MOD(number,2)
number = number/2
ENDDO
! write out results
WRITE(*,*) 'Binary representation '
DO i=0, 31
WRITE(*,*)' Term nr and value', i, terms(i)
ENDDO
END PROGRAM binary_integer
```

**Possible Overflow for Integers.**

```
// A comment line begins like this in C++ programs
// Program to calculate 2**n
// Standard ANSI-C++ include files */
#include <iostream>
#include <cmath>
using namespace std
int main()
{
int int1, int2, int3;
// print to screen
cout << "Read in the exponential N for 2^N =" << endl;
// read from screen
cin >> int2;
int1 = (int) pow(2., (double) int2);
cout << " 2^N * 2^N = " << int1*int1 << endl;
int3 = int1 - 1;
cout << " 2^N*(2^N - 1) = " << int1 * int3 << endl;
cout << " 2^N- 1 = " << int3 << endl;
return 0;
// End: program main()
```

$$
9.90625=0.990625\times 10^{1},
$$

and a real non-zero number could be generalized as

$$
\begin{equation}
x=\pm r\times 10^{{\mbox{n}}},
\label{_auto1} \tag{1}
\end{equation}
$$

$$
\begin{equation}
x=\pm q\times 2^{{\mbox{m}}},
\label{_auto2} \tag{2}
\end{equation}
$$

$$
\begin{equation}
(0.a_{-1}a_{-2}\dots a_{-n})_2=a_{-1}\times 2^{-1}
+a_{-2}\times 2^{-2}+\dots+a_{-n}\times 2^{-n}.
\label{_auto3} \tag{3}
\end{equation}
$$

$$
\begin{equation}
x=(-1)^s\times {\mbox{mantissa}}\times 2^{{\mbox{exponent}}},
\label{_auto4} \tag{4}
\end{equation}
$$

where $s$ is the sign bit, and the exponent gives the available range. With a single-precision word, 32 bits, 8 bits would typically be reserved for the exponent, 1 bit for the sign and 23 for the mantissa.

**Machine Numbers.**

A modification of the scientific notation for binary numbers is to require that the leading binary digit 1 appears to the left of the binary point. In this case the representation of the mantissa $q$ would be $(1.f)_2$ and $ 1 \le q < 2$. This form is rather useful when storing binary numbers in a computer word, since we can always assume that the leading bit 1 is there. One bit of space can then be saved meaning that a 23 bits mantissa has actually 24 bits. This means explicitely that a binary number with 23 bits for the mantissa reads

$$
\begin{equation}
(1.a_{-1}a_{-2}\dots a_{-23})_2=1\times 2^0+a_{-1}\times 2^{-1}+
+a_{-2}\times 2^{-2}+\dots+a_{-23}\times 2^{-23}.
\label{_auto5} \tag{5}
\end{equation}
$$

$$
(10111110111101000000000000000000)_2,
$$

where the first bit is reserved for the sign, 1 in this case yielding a negative sign. The exponent $m$ is given by the next 8 binary numbers $01111101$ resulting in 125 in the decimal system.

**Machine Numbers.**

However, since the exponent has eight bits, this means it has $2^8-1=255$ possible numbers in the interval $-128 \le m \le 127$, our final exponent is $125-127=-2$ resulting in $2^{-2}$. Inserting the sign and the mantissa yields the final number in the decimal representation as

$$
-2^{-2}\left(1\times 2^0+1\times 2^{-1}+
1\times 2^{-2}+1\times 2^{-3}+0\times 2^{-4}+1\times 2^{-5}\right)=
$$

$$
(-0.4765625)_{10}.
$$

In this case we have an exact machine representation with 32 bits (actually, we need less than 23 bits for the mantissa).

If our number $x$ can be exactly represented in the machine, we call $x$ a machine number. Unfortunately, most numbers cannot and are thereby only approximated in the machine. When such a number occurs as the result of reading some input data or of a computation, an inevitable error will arise in representing it as accurately as possible by a machine number.

**Machine Numbers.**

A floating number x, labelled $fl(x)$ will therefore always be represented as

$$
\begin{equation}
fl(x) = x(1\pm \epsilon_x),
\label{_auto6} \tag{6}
\end{equation}
$$

$$
\left(1.a_{-1}a_{-2}\dots a_{-n}\right)_2
$$

is always truncated at some stage $n$ due to its limited number of bits, there is only a limited number of real binary numbers. The spacing between every real binary number is given by the chosen machine precision. For a 32 bit words this number is approximately $ \epsilon_M \sim 10^{-7}$ and for double precision (64 bits) we have $ \epsilon_M \sim 10^{-16}$, or in terms of a binary base as $2^{-23}$ and $2^{-52}$ for single and double precision, respectively.

**Machine Numbers.**

In the machine a number is represented as

$$
\begin{equation}
fl(x)= x(1+\epsilon)
\label{_auto7} \tag{7}
\end{equation}
$$

$$
\begin{equation}
fl(a)=fl(b)-fl(c) = a(1+\epsilon_a),
\label{_auto8} \tag{8}
\end{equation}
$$

or

$$
\begin{equation}
fl(a)=b(1+\epsilon_b)-c(1+\epsilon_c),
\label{_auto9} \tag{9}
\end{equation}
$$

$$
\begin{equation}
fl(a)/a=1+\epsilon_b\frac{b}{a}- \epsilon_c\frac{c}{a},
\label{_auto10} \tag{10}
\end{equation}
$$

$$
\begin{equation}
|fl(a)-a|,
\label{_auto11} \tag{11}
\end{equation}
$$

whereas the relative error is

$$
\begin{equation}
\frac{ |fl(a)-a|}{a} \le \epsilon_a.
\label{_auto12} \tag{12}
\end{equation}
$$

$$
\begin{equation}
\frac{ |fl(a)-a|}{a}=\frac{ |fl(b)-fl(c)-(b-c)|}{a},
\label{_auto13} \tag{13}
\end{equation}
$$

yielding

$$
\begin{equation}
\frac{ |fl(a)-a|}{a}=\frac{ |b\epsilon_b- c\epsilon_c|}{a}.
\label{_auto14} \tag{14}
\end{equation}
$$

$$
f(x)=\frac{1-\cos(x)}{\sin(x)},
$$

$$
f(x)=\frac{\sin(x)}{1+\cos(x)}.
$$

If we now choose $x=0.007$ (in radians) our choice of precision results in

$$
\sin(0.007)\approx 0.69999\times 10^{-2},
$$

and

$$
\cos(0.007)\approx 0.99998.
$$

$$
f(x)=\frac{1-0.99998}{0.69999\times 10^{-2}}=\frac{0.2\times 10^{-4}}{0.69999\times 10^{-2}}=0.28572\times 10^{-2},
$$

while the second expression results in

$$
f(x)=\frac{0.69999\times 10^{-2}}{1+0.99998}=
\frac{0.69999\times 10^{-2}}{1.99998}=0.35000\times 10^{-2},
$$

which is also the exact result. In the first expression, due to our choice of precision, we have only one relevant digit in the numerator, after the subtraction. This leads to a loss of precision and a wrong result due to a cancellation of two nearly equal numbers. If we had chosen a precision of six leading digits, both expressions yield the same answer.

If we were to evaluate $x\sim \pi$, then the second expression for $f(x)$ can lead to potential losses of precision due to cancellations of nearly equal numbers.

This simple example demonstrates the loss of numerical precision due
to roundoff errors, where the number of leading digits is lost
in a subtraction of two near equal numbers.
The lesson to be drawn is that we cannot blindly compute a function.
We will always need to carefully analyze our algorithm in the search for
potential pitfalls. There is no magic recipe however, the only guideline
is an understanding of the fact that a machine cannot represent
correctly *all* numbers.

**Real Numbers.**

**Overflow**: When the positive exponent exceeds the max value, e.g., 308 for`DOUBLE PRECISION`

(64 bits). Under such circumstances the program will terminate and some compilers may give you the warning`OVERFLOW`

.**Underflow**: When the negative exponent becomes smaller than the min value, e.g., -308 for`DOUBLE PRECISION`

. Normally, the variable is then set to zero and the program continues. Other compilers (or compiler options) may warn you with the`UNDERFLOW`

message and the program terminates.

A floating point number like

$$
\begin{equation}
x= 1.234567891112131468 = 0.1234567891112131468\times 10^{1}
\label{_auto15} \tag{15}
\end{equation}
$$

may be stored in the following way. The exponent is small and is stored in full precision. However, the mantissa is not stored fully. In double precision (64 bits), digits beyond the 15th are lost since the mantissa is normally stored in two words, one which is the most significant one representing 123456 and the least significant one containing 789111213. The digits beyond 3 are lost. Clearly, if we are summing alternating series with large numbers, subtractions between two large numbers may lead to roundoff errors, since not all relevant digits are kept. This leads eventually to the next problem, namely

**Real Numbers.**

**Loss of precision**: When one has to e.g., multiply two large numbers where one suspects that the outcome may be beyond the bonds imposed by the variable declaration, one could represent the numbers by logarithms, or rewrite the equations to be solved in terms of dimensionless variables. When dealing with problems in e.g., particle physics or nuclear physics where distance is measured in fm ($10^{-15}$ m), it can be quite convenient to redefine the variables for distance in terms of a dimensionless variable of the order of unity. To give an example, suppose you work with single precision and wish to perform the addition $1+10^{-8}$. In this case, the information containing in $10^{-8}$ is simply lost in the addition. Typically, when performing the addition, the computer equates first the exponents of the two numbers to be added. For $10^{-8}$ this has however catastrophic consequences since in order to obtain an exponent equal to $10^0$, bits in the mantissa are shifted to the right. At the end, all bits in the mantissa are zeros.

**Three ways of computing $e^{-x}$.**

Brute force:

$$
\exp{(-x)}=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}
$$

Recursion relation for

$$
\exp{(-x)}=\sum_{n=0}^{\infty}s_n=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}
$$

$$
s_n=-s_{n-1}\frac{x}{n},
$$

$$
\exp{(x)}=\sum_{n=0}^{\infty}s_n
$$

$$
\exp{(-x)}=\frac{1}{\exp{(x)}}
$$

**Brute Force.**

```
// Program to calculate function exp(-x)
// using straightforward summation with differing precision
using namespace std
#include <iostream>
#include <cmath>
// type float: 32 bits precision
// type double: 64 bits precision
#define TYPE double
#define PHASE(a) (1 - 2 * (abs(a) % 2))
#define TRUNCATION 1.0E-10
// function declaration
TYPE factorial(int);
```

```
int main()
{
int n;
TYPE x, term, sum;
for(x = 0.0; x < 100.0; x += 10.0) {
sum = 0.0; //initialization
n = 0;
term = 1;
while(fabs(term) > TRUNCATION) {
term = PHASE(n) * (TYPE) pow((TYPE) x,(TYPE) n)
/ factorial(n);
sum += term;
n++;
} // end of while() loop
```

```
printf("\nx = %4.1f exp = %12.5E series = %12.5E
number of terms = %d",
x, exp(-x), sum, n);
} // end of for() loop
printf("\n"); // a final line shift on output
return 0;
} // End: function main()
// The function factorial()
// calculates and returns n!
TYPE factorial(int n)
{
int loop;
TYPE fac;
for(loop = 1, fac = 1.0; loop <= n; loop++) {
fac *= loop;
return fac;
} // End: function factorial()
```

**What is going on?**

$x$ | $\exp{(-x)}$ | Series | Number of terms in series |
---|---|---|---|

0.0 | 0.100000E+01 | 0.100000E+01 | 1 |

10.0 | 0.453999E-04 | 0.453999E-04 | 44 |

20.0 | 0.206115E-08 | 0.487460E-08 | 72 |

30.0 | 0.935762E-13 | -0.342134E-04 | 100 |

40.0 | 0.424835E-17 | -0.221033E+01 | 127 |

50.0 | 0.192875E-21 | -0.833851E+05 | 155 |

60.0 | 0.875651E-26 | -0.850381E+09 | 171 |

70.0 | 0.397545E-30 | NaN | 171 |

80.0 | 0.180485E-34 | NaN | 171 |

90.0 | 0.819401E-39 | NaN | 171 |

100.0 | 0.372008E-43 | NaN | 171 |

```
// program to compute exp(-x) without exponentials
using namespace std
#include <iostream>
#include <cmath>
#define TRUNCATION 1.0E-10
int main()
{
int loop, n;
double x, term, sum;
for(loop = 0; loop <= 100; loop += 10)
{
x = (double) loop; // initialization
sum = 1.0;
term = 1;
n = 1;
```

```
while(fabs(term) > TRUNCATION)
{
term *= -x/((double) n);
sum += term;
n++;
} // end while loop
cout << "x = " << x << " exp = " << exp(-x) <<"series = "
<< sum << " number of terms =" << n << endl;
} // end of for() loop
cout << endl; // a final line shift on output
} /* End: function main() */
```

**More Problems.**

$x$ | $\exp{(-x)}$ | Series | Number of terms in series |
---|---|---|---|

0.000000 | 0.10000000E+01 | 0.10000000E+01 | 1 |

10.000000 | 0.45399900E-04 | 0.45399900E-04 | 44 |

20.000000 | 0.20611536E-08 | 0.56385075E-08 | 72 |

30.000000 | 0.93576230E-13 | -0.30668111E-04 | 100 |

40.000000 | 0.42483543E-17 | -0.31657319E+01 | 127 |

50.000000 | 0.19287498E-21 | 0.11072933E+05 | 155 |

60.000000 | 0.87565108E-26 | -0.33516811E+09 | 182 |

70.000000 | 0.39754497E-30 | -0.32979605E+14 | 209 |

80.000000 | 0.18048514E-34 | 0.91805682E+17 | 237 |

90.000000 | 0.81940126E-39 | -0.50516254E+22 | 264 |

100.000000 | 0.37200760E-43 | -0.29137556E+26 | 291 |

**3 point formulae.**

First derivative ($f_0 = f(x_0)$, $f_{-h}=f(x_0-h)$ and $f_{+h}=f(x_0+h)$

$$
\frac{f_h-f_{-h}}{2h}=f'_0+\sum_{j=1}^{\infty}\frac{f_0^{(2j+1)}}{(2j+1)!}h^{2j}.
$$

Second derivative

$$
\frac{ f_h -2f_0 +f_{-h}}{h^2}=f_0''+2\sum_{j=1}^{\infty}\frac{f_0^{(2j+2)}}{(2j+2)!}h^{2j}.
$$

$$
\epsilon=log_{10}\left(\left|\frac{f''_{\mbox{computed}}-f''_{\mbox{exact}}}
{f''_{\mbox{exact}}}\right|\right),
$$

$$
\epsilon_{\mbox{tot}}=\epsilon_{\mbox{approx}}+\epsilon_{\mbox{ro}}.
$$

For the computed second derivative we have

$$
f_0''=\frac{ f_h -2f_0 +f_{-h}}{h^2}-2\sum_{j=1}^{\infty}\frac{f_0^{(2j+2)}}{(2j+2)!}h^{2j},
$$

and the truncation or approximation error goes like

$$
\epsilon_{\mbox{approx}}\approx \frac{f_0^{(4)}}{12}h^{2}.
$$

$$
f_0''=\frac{ f_h -2f_0 +f_{-h}}{h^2}=\frac{ (f_h -f_0) +(f_{-h}-f_0)}{h^2},
$$

we reach fairly quickly a limit for where loss of precision due to the subtraction of two nearly equal numbers becomes crucial.

If $(f_{\pm h} -f_0)$ are very close, we have $(f_{\pm h} -f_0)\approx \epsilon_M$, where $|\epsilon_M|\le 10^{-7}$ for single and $|\epsilon_M|\le 10^{-15}$ for double precision, respectively.

We have then

$$
\left|f_0''\right|=
\left|\frac{ (f_h -f_0) +(f_{-h}-f_0)}{h^2}\right|\le \frac{ 2 \epsilon_M}{h^2}.
$$

$$
\left|\epsilon_{\mbox{tot}}\right|\le \frac{2 \epsilon_M}{h^2} +
\frac{f_0^{(4)}}{12}h^{2}.
\label{eq:experror} \tag{16}
$$

$$
h= \left(\frac{ 24\epsilon_M}{f_0^{(4)}}\right)^{1/4}.
$$

With double precision and $x=10$ we obtain

$$
h\approx 10^{-4}.
$$

Beyond this value, it is essentially the loss of numerical precision which takes over.

Due to the subtractive cancellation in the expression for $f''$ there is a pronounced detoriation in accuracy as $h$ is made smaller and smaller.

It is instructive in this analysis to rewrite the numerator of the computed derivative as

$$
(f_h -f_0) +(f_{-h}-f_0)=(e^{x+h}-e^{x}) + (e^{x-h}-e^{x}),
$$

as

$$
(f_h -f_0) +(f_{-h}-f_0)=e^x(e^{h}+e^{-h}-2),
$$

since it is the difference $(e^{h}+e^{-h}-2)$ which causes the loss of precision.

$x$ | $h=0.01$ | $h=0.001$ | $h=0.0001$ | $h=0.0000001$ | Exact |
---|---|---|---|---|---|

0.0 | 1.000008 | 1.000000 | 1.000000 | 1.010303 | 1.000000 |

1.0 | 2.718304 | 2.718282 | 2.718282 | 2.753353 | 2.718282 |

2.0 | 7.389118 | 7.389057 | 7.389056 | 7.283063 | 7.389056 |

3.0 | 20.085704 | 20.085539 | 20.085537 | 20.250467 | 20.085537 |

4.0 | 54.598605 | 54.598155 | 54.598151 | 54.711789 | 54.598150 |

5.0 | 148.414396 | 148.413172 | 148.413161 | 150.635056 | 148.413159 |

The results for $x=10$ are shown in the Table

$h$ | $e^{h}+e^{-h}$ | $e^{h}+e^{-h}-2$ |
---|---|---|

$10^{-1}$ | 2.0100083361116070 | $1.0008336111607230\times 10^{-2}$ |

$10^{-2}$ | 2.0001000008333358 | $1.0000083333605581\times 10^{-4}$ |

$10^{-3}$ | 2.0000010000000836 | $1.0000000834065048\times 10^{-6}$ |

$10^{-5}$ | 2.0000000099999999 | $1.0000000050247593\times 10^{-8}$ |

$10^{-5}$ | 2.0000000001000000 | $9.9999897251734637\times 10^{-11}$ |

$10^{-6}$ | 2.0000000000010001 | $9.9997787827987850\times 10^{-13}$ |

$10^{-7}$ | 2.0000000000000098 | $9.9920072216264089\times 10^{-15}$ |

$10^{-8}$ | 2.0000000000000000 | $0.0000000000000000\times 10^{0}$ |

$10^{-9}$ | 2.0000000000000000 | $1.1102230246251565\times 10^{-16}$ |

$10^{-10}$ | 2.0000000000000000 | $0.0000000000000000\times 10^{0}$ |

We list here the program to compute the above derivative

```
#include <iostream>
#include <cmath>
#include <fstream>
#include <iomanip>
// Note: not using namespace for std
// output file as global variable
std::ofstream ofile;
// Begin of main program
int main(int argc, char* argv[])
{
char *outfilename;
// Read in output file, abort if there are too few command-line arguments
if( argc <= 3 ){
std::cout << "Bad Usage: " << argv[0] <<
" read also output file and number of elements on same line" << std::endl;
exit(1);
}
else{
outfilename=argv[1];
}
// opening a file for the program
ofile.open(outfilename);
// extracting number of mesh points
int i = atoi(argv[2]);
double x = atof(argv[3]); // reading x-value
double h = 1.0/((double) i); // setting up step size
double Derivative = (exp(x+h)-2.*exp(x)+exp(x-h))/(h*h);
double RelativeError = log10(fabs(Derivative-exp(x))/exp(x));
ofile << std::setw(15) << std::setprecision(8) << "relative error=" << RelativeError << std::endl;
ofile.close(); // close output file
return 0;
}
```

```
using namespace std; // note use of namespace
int main()
{
// what are the differences?
int var;
cin >> var;
int *p, q;
int *s, *t;
int * a new[var]; // dynamic memory allocation
delete [] a;
}
```

```
using namespace std; // note use of namespace
int main()
{
int var;
int *p;
p = &var
var = 421;
printf("Address of integer variable var : %p\n",&var);
printf("Its value: %d\n", var);
printf("Value of integer pointer p : %p\n",p);
printf("The value p points at : %d\n",*p);
printf("Address of the pointer p : %p\n",&p);
return 0;
}
```

```
int main()
{
int var; // Define an integer variable var
int *p; // Define a pointer to an integer
p = &var // Extract the address of var
var = 421; // Change content of var
printf("Address of integer variable var : %p\n", &var);
printf("Its value: %d\n", var); // 421
printf("Value of integer pointer p : %p\n", p); // = &var
// The content of the variable pointed to by p is *p
printf("The value p points at : %d\n", *p);
// Address where the pointer is stored in memory
printf("Address of the pointer p : %p\n", &p);
return 0;
}
```

```
int matr[2];
int *p;
p = &matr[0];
matr[0] = 321;
matr[1] = 322;
printf("\nAddress of matrix element matr[1]: %p",&matr[0]);
printf("\nValue of the matrix element matr[1]; %d",matr[0]);
printf("\nAddress of matrix element matr[2]: %p",&matr[1]);
printf("\nValue of the matrix element matr[2]: %d\n", matr[1]);
printf("\nValue of the pointer p: %p",p);
printf("\nThe value p points to: %d",*p);
printf("\nThe value that (p+1) points to %d\n",*(p+1));
printf("\nAddress of pointer p : %p\n",&p);
```

```
int matr[2]; // Define integer array with two elements
int *p; // Define pointer to integer
p = &matr[0]; // Point to the address of the first element in matr
matr[0] = 321; // Change the first element
matr[1] = 322; // Change the second element
printf("\nAddress of matrix element matr[1]: %p", &matr[0]);
printf("\nValue of the matrix element matr[1]; %d", matr[0]);
printf("\nAddress of matrix element matr[2]: %p", &matr[1]);
printf("\nValue of the matrix element matr[2]: %d\n", matr[1]);
printf("\nValue of the pointer p: %p", p);
printf("\nThe value p points to: %d", *p);
printf("\nThe value that (p+1) points to %d\n", *(p+1));
printf("\nAddress of pointer p : %p\n", &p);
```

```
Address of the matrix element matr[1]: 0xbfffef70
Value of the matrix element matr[1]; 321
Address of the matrix element matr[2]: 0xbfffef74
Value of the matrix element matr[2]: 322
Value of the pointer: 0xbfffef70
The value pointer points at: 321
The value that (pointer+1) points at: 322
Address of the pointer variable : 0xbfffef6c
```

```
using namespace std;
#include <iostream>
int main(int argc, char *argv[])
{
FILE *in_file, *out_file;
if( argc < 3) {
printf("The programs has the following structure :\n");
printf("write in the name of the input and output files \n");
exit(0);
}
in_file = fopen( argv[1], "r");// returns pointer to the input file
if( in_file == NULL ) { // NULL means that the file is missing
printf("Can't find the input file %s\n", argv[1]);
exit(0);
```

```
out_file = fopen( argv[2], "w"); // returns a pointer to the output file
if( out_file == NULL ) { // can't find the file
printf("Can't find the output file%s\n", argv[2]);
exit(0);
}
fclose(in_file);
fclose(out_file);
return 0;
```

```
#include <fstream>
// input and output file as global variable
ofstream ofile;
ifstream ifile;
```

```
int main(int argc, char* argv[])
{
char *outfilename;
//Read in output file, abort if there are too
//few command-line arguments
if( argc <= 1 ){
cout << "Bad Usage: " << argv[0] <<
" read also output file on same line" << endl;
exit(1);
}
else{
outfilename=argv[1];
}
ofile.open(outfilename);
.....
ofile.close(); // close output file
```

```
void output(double r_min , double r_max, int max_step,
double *d)
{
int i;
ofile << "RESULTS:" << endl;
ofile << setiosflags(ios::showpoint | ios::uppercase);
ofile <<"R_min = " << setw(15) << setprecision(8) <<r_min <<endl;
ofile <<"R_max = " << setw(15) << setprecision(8) <<r_max <<endl;
ofile <<"Number of steps = " << setw(15) << max_step << endl;
ofile << "Five lowest eigenvalues:" << endl;
for(i = 0; i < 5; i++) {
ofile << setw(15) << setprecision(8) << d[i] << endl;
} // end of function output
```

```
int main(int argc, char* argv[])
{
char *infilename;
// Read in input file, abort if there are too
// few command-line arguments
if( argc <= 1 ){
cout << "Bad Usage: " << argv[0] <<
" read also input file on same line" << endl;
exit(1);
}
else{
infilename=argv[1];
}
ifile.open(infilename);
....
ifile.close(); // close input file
```

```
const char* filename1 = "myfile";
ifstream ifile(filename1);
string filename2 = filename1 + ".out"
ofstream ofile(filename2); // new output file
ofstream ofile(filename2, ios_base::app); // append
// Read something from the file:
double a; int b; char c[200];
ifile >> a >> b >> c; // skips white space in between
// Can test on success of reading:
if (!(ifile >> a >> b >> c)) ok = 0;
```

```
int n; n =8;
func(&n); /* &n is a pointer to n */
....
void func(int *i)
{
*i = 10; /* n is changed to 10 */
....
}
```

whereas in C++ we would write

```
int n; n =8;
func(n); // just transfer n itself
....
void func(int& i)
{
i = 10; // n is changed to 10
....
}
```

The reason why we emphasize the difference between call by value and call by reference is that it allows the programmer to avoid pitfalls like unwanted changes of variables. However, many people feel that this reduces the readability of the code.

In Fortran we can use `INTENT(IN)`

, `INTENT(OUT)`

, `INTENT(INOUT)`

to let the
program know which values should or should not be changed.

```
SUBROUTINE coulomb_integral(np,lp,n,l,coulomb)
USE effective_interaction_declar
USE energy_variables
USE wave_functions
IMPLICIT NONE
INTEGER, INTENT(IN) :: n, l, np, lp
INTEGER :: i
REAL(KIND=8), INTENT(INOUT) :: coulomb
REAL(KIND=8) :: z_rel, oscl_r, sum_coulomb
...
```

This hinders unwanted changes and increases readability.

```
#include <iostream>
#include <cmath>
using namespace std; // note use of namespace
int main (int argc, char* argv[])
{
int i = atoi(argv[1]);
// Dynamic memory allocation: need tp declare -a- as a pointer
// You can use double *a = new double[i]; or
double *a;
a = new double[i];
// the first of element of a, a[0], and its address is the
// value of the pointer.
/* This is a longer comment
if we want a static memory allocation
this is the way to do it
*/
cout << " bytes for i=" << sizeof(i) << endl;
for (int j = 0; j < i; j++) {
a[j] = j*exp(2.0);
cout << "a=" << a[j] << endl;
}
// freeing memory
delete [] a;
// to check for memory leaks, use the software called -valgrind-
return 0; /* success execution of the program */
}
```

```
#include <iostream>
#include <cmath>
#include <fstream>
#include <iomanip>
using namespace std; // note use of namespace
// output file as global variable
ofstream ofile;
// Begin of main program
int main(int argc, char* argv[])
{
char *outfilename;
// Read in output file, abort if there are too few command-line arguments
if( argc <= 2 ){
cout << "Bad Usage: " << argv[0] <<
" read also output file and number of elements on same line" << endl;
exit(1);
}
else{
outfilename=argv[1];
}
// opening a file for the program
ofile.open(outfilename);
int i = atoi(argv[2]);
// int *a;
//a = new int[i];
double *a = new double[i];
cout << " bytes for i=" << sizeof(i) << endl;
for (int j = 0; j < i; j++) {
a[j] = j*exp(2.0);
// ofile instead of cout
ofile << setw(15) << setprecision(8) << "a=" << a[j] << endl;
}
delete [] a; // free memory
ofile.close(); // close output file
return 0;
}
```

```
#include <iostream>
using namespace std;
// Declare functions before main
void func(int, int*);
int main(int argc, char *argv[])
{
int a;
int *b;
a = 10;
b = new int[10];
for(int i = 0; i < 10; i++) {
b[i] = i;
cout << b[i] << endl;
}
// the variable a is transferred by call by value. This means
// that the function func cannot change a in the calling function
func( a,b);
delete [] b ;
return 0;
} // End: function main()
void func( int x, int *y)
{
// a becomes locally x and it can be changed locally
x+=7;
// func gets the address of the first element of y (b)
// it changes y[0] to 10 and when returning control to main
// it changes also b[0]. Call by reference
*y += 10; // *y = *y+10;
// explicit element
y[6] += 10;
// in this function y[0] and y[6] have been changed and when returning
// control to main this means that b[0] and b[6] are changed.
return;
} // End: function func()
```

```
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <iomanip>
#include "time.h"
using namespace std; // note use of namespace
int main (int argc, char* argv[])
{
int i = atoi(argv[1]);
double *a, *b, *c;
a = new double[i];
b = new double[i];
c = new double[i];
clock_t start, finish;
start = clock();
for (int j = 0; j < i; j++) {
a[j] = cos(j*1.0);
b[j] = sin(j+3.0);
c[j] = 0.0;
}
for (int j = 0; j < i; j++) {
c[j] = a[j]+b[j];
}
finish = clock();
double timeused = (double) (finish - start)/(CLOCKS_PER_SEC );
cout << setiosflags(ios::showpoint | ios::uppercase);
cout << setprecision(10) << setw(20) << "Time used for vector addition=" << timeused << endl;
delete [] a;
delete [] b;
delete [] c;
return 0; /* success execution of the program */
}
```