The problem is to find the real root of the equation: f(x)=x^(x^8)-8=0. I use the Newton-Raphson method, a root finding algorithm. A first guess is x = 2. The solution is: x = 1.2968395547, f(x) = -2.6645353e-15. I compute the necessary derivative using central-finite differences with a step size of h = 2/10000.
Category: Memoirs of James Pate Williams Jr
Blog Entry August 1, 2024, (c) James Pate Williams, Jr. Online Math Problem
Blog Entry Thursday, July 25, 2024, (c) James Pate Williams, Jr. Newton’s Method for Finding the Real Roots of a Real Polynomial
degree (0 to quit) = 5
a[5] = 1
a[4] = -15.5
a[3] = 77.5
a[2] = -155
a[1] = 124
a[0] = -32
x[0] = 0.45
x[1] = 0.9
x[2] = 1.8
x[3] = 3.6
x[4] = 7.2
iterations = 31
root[0] = 5.0000000000e-01
root[1] = 1.0000000000e+00
root[2] = 2.0000000000e+00
root[3] = 4.0000000000e+00
root[4] = 8.0000000000e+00
func[0] = 0.0000000000e+00
func[1] = 0.0000000000e+00
func[2] = 0.0000000000e+00
func[3] = 0.0000000000e+00
func[4] = 0.0000000000e+00
degree (0 to quit) =
Blog Entry (c) Tuesday, July 23, 2024, by James Pate Williams, Jr. Mueller’s Method for Finding the Complex and/or Real Roots of a Complex and/or Real Polynomial
I originally implemented this algorithm in FORTRAN IV in the Summer Quarter of 1982 at the Georgia Institute of Technology. I was taking a course named “Scientific Computing I” taught by Professor Gunter Meyer. I made a B in the class. Later in 2015 I re-implemented the recipe in C# using Visual Studio 2008 Professional. VS 2015 did not have support for complex numbers nor large integers. In December of 2015 I upgraded to Visual Studio 2015 Professional which has support for big integers and complex numbers. I used Visual Studio 2019 Community version for this project. Root below should be function.
Degree (0 to quit) = 2
coefficient[2].real = 1
coefficient[2].imag = 0
coefficient[1].real = 1
coefficient[1].imag = 0
coefficient[0].real = 1
coefficient[0].imag = 0
zero[0].real = -5.0000000000e-01 zero[0].imag = 8.6602540378e-01
zero[1].real = -5.0000000000e-01 zero[1].imag = -8.6602540378e-01
root[0].real = 0.0000000000e+00 root[0].imag = -2.2204460493e-16
root[1].real = 3.3306690739e-16 root[1].imag = -7.7715611724e-16
Degree (0 to quit) = 3
coefficient[3].real = 1
coefficient[3].imag = 0
coefficient[2].real = 0
coefficient[2].imag = 0
coefficient[1].real = -18.1
coefficient[1].imag = 0
coefficient[0].real = -34.8
coefficient[0].imag = 0
zero[0].real = -2.5026325486e+00 zero[0].imag = -8.3036679880e-01
zero[1].real = -2.5026325486e+00 zero[1].imag = 8.3036679880e-01
zero[2].real = 5.0052650973e+00 zero[2].imag = 2.7417672687e-15
root[0].real = 0.0000000000e+00 root[0].imag = 1.7763568394e-15
root[1].real = 3.5527136788e-14 root[1].imag = -1.7763568394e-14
root[2].real = 2.8421709430e-14 root[2].imag = 1.5643985575e-13
Degree (0 to quit) = 5
coefficient[5].real = 1
coefficient[5].imag = 0
coefficient[4].real = 2
coefficient[4].imag = 0
coefficient[3].real = 3
coefficient[3].imag = 0
coefficient[2].real = 4
coefficient[2].imag = 0
coefficient[1].real = 5
coefficient[1].imag = 0
coefficient[0].real = 6
coefficient[0].imag = 0
zero[0].real = -8.0578646939e-01 zero[0].imag = 1.2229047134e+00
zero[1].real = -8.0578646939e-01 zero[1].imag = -1.2229047134e+00
zero[2].real = 5.5168546346e-01 zero[2].imag = 1.2533488603e+00
zero[3].real = 5.5168546346e-01 zero[3].imag = -1.2533488603e+00
zero[4].real = -1.4917979881e+00 zero[4].imag = 1.8329656063e-15
root[0].real = 8.8817841970e-16 root[0].imag = 4.4408920985e-16
root[1].real = -2.6645352591e-15 root[1].imag = -4.4408920985e-16
root[2].real = 8.8817841970e-16 root[2].imag = 1.7763568394e-15
root[3].real = 3.4638958368e-14 root[3].imag = -1.4210854715e-14
root[4].real = 8.8817841970e-16 root[4].imag = 2.0710031449e-14
Blog Entry Sunday, July 21, 2024 (c) James Pate Williams, Jr. Another Easy Internet Mathematics Problem
x = 1.2679491924e+00 y = 4.7320508076e+00
f = 0.0000000000e+00 g = 0.0000000000e+00
iterations = 100
legend: f = x + y – 6, g = x * y – 6
The solution was found via my Win32 C application whose source code is presented below (the method is the Newton iteration for systems of linear and/or non-linear equations):
Blog Entry Friday, July 19, 2024, Easy Internet Math “Puzzle” (c) James Pate Williams, Jr.
#include <math.h>
#include <iostream>
using namespace std;
long double f(long double x)
{
return powl(8.0, x) - powl(2.0, x) -
2.0 * (powl(6.0, x) - powl(3.0, x));
}
long double g(long double x)
{
return powl(8.0, x) * logl(8.0) - powl(2.0, x) * logl(2.0) -
2.0 * (powl(6.0, x) * logl(6.0) - powl(3.0, x) * logl(3.0));
}
long double Newton(long double x, int maxIts, int& iterations)
{
long double x0 = x;
long double x1 = 0.0;
iterations = 0;
while (true) {
long double dx = 0.0;
long double fx = f(x0);
long double gx = g(x0);
x1 = x0 - fx / gx;
dx = fabsl(x1 - x0);
iterations++;
if (dx < 1.0e-15)
break;
if (fabsl(fx) < 1.0e-15)
break;
if (iterations == maxIts)
break;
x0 = x1;
}
return x1;
}
int main() {
int iterations = 0, maxIts;
long double x0 = 0.0, x1 = 0.0;
while (true) {
cout << "x0 = ";
cin >> x0;
if (x0 == 0)
break;
cout << "maximum iterations = ";
cin >> maxIts;
x1 = Newton(x0, maxIts, iterations);
cout << "x1 = " << x1 << endl;
cout << "iterations = ";
cout << iterations << endl;
}
return 0;
}
Blog Entry Tuesday, July 16, 2024, Prime Number Zeta Function Values for s = 2 and Primes < 1,000,000 (c) James Pate Williams, Jr.
Runtime in seconds to build sieve: 4.462000
N = <= 1000000 17
s > 1 = 2
prime zeta = 0.449282078106864
N = <= 1000000 257
s > 1 = 2
prime zeta = 0.452175428603043
N = <= 1000000 65537
s > 1 = 2
prime zeta = 0.452247336475390
N = <= 1000000 100003
s > 1 = 2
prime zeta = 0.452247368830138
N = <= 1000000 900001
s > 1 = 2
prime zeta = 0.452247415884861
N = <= 1000000 0
C:\Users\james\source\repos\PrimeZetaFunction\Debug\PrimeZetaFunction.exe (process 20584) exited with code 0.
Press any key to close this window . . .
/*
* PrimeZetaFunction.c (c) Monday, July 15, 2024
* by James Pate Williams, Jr.
*/
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define BITS_PER_LONG 32
#define BITS_PER_LONG_1 31
#define MAX_SIEVE 100000000
#define MAX_PRIME_INDEX 5761454
#define SIEVE_SIZE (MAX_SIEVE / BITS_PER_LONG + 1)
long prime[MAX_PRIME_INDEX], sieve[SIEVE_SIZE];
long get_bit(long i, long* sieve)
{
long b = i % BITS_PER_LONG;
long c = i / BITS_PER_LONG;
return (sieve[c] >> (BITS_PER_LONG_1 - b)) & 1;
}
void set_bit(long i, long v, long* sieve)
{
long b = i % BITS_PER_LONG;
long c = i / BITS_PER_LONG;
long mask = 1 << (BITS_PER_LONG_1 - b);
if (v == 1)
sieve[c] |= mask;
else
sieve[c] &= ~mask;
}
void Sieve(long n, long* sieve)
{
long c, i, inc;
set_bit(0, 0, sieve);
set_bit(1, 0, sieve);
set_bit(2, 1, sieve);
for (i = 3; i <= n; i++)
set_bit(i, i & 1, sieve);
c = 3;
do {
i = c * c, inc = c + c;
while (i <= n) {
set_bit(i, 0, sieve);
i += inc;
}
c += 2;
while (!get_bit(c, sieve)) c++;
} while (c * c <= n);
}
long double primeZetaFunction(long N, long s)
{
long n, p;
long double sum = 0.0L;
for (n = N; n >= 0; n--)
{
p = prime[n];
sum += 1.0 / powl((long double)p, (long double)s);
}
return sum;
}
int main()
{
long N = 0, i = 0, p = 2, s = 0;
double runtime = 0.0;
clock_t time0 = clock(), time1 = 0;
Sieve(MAX_SIEVE, sieve);
for (i = 0; i <= MAX_PRIME_INDEX; i++) {
while (!get_bit(p, sieve)) p++;
prime[i] = p++;
}
time1 = clock();
runtime = ((double)time1 - time0) / CLOCKS_PER_SEC;
printf_s("Runtime in seconds to build sieve: %Lf\n", runtime);
for (;;) {
printf_s("N = <= %ld ", 1000000);
scanf_s("%ld", &N);
if (N == 0)
break;
printf_s("s > 1 = ");
scanf_s("%ld", &s);
printf_s("prime zeta = %16.15Lf\n", primeZetaFunction(N, s));
}
return 0;
}
Blog Entry Monday, July 15, 2024, (c) James Pate Williams, Jr. Some Values of the Riemann Zeta Function Computed with Fast and Very Slow Algorithms
The slow computations used 999,999,999 terms. I seem to recall from my first numerical analysis (Scientific Computing I) course in the Mathematics Department at the Georgia Institute of Technology with Professor Gunter Meyer in the Summer of 1982 that computing a truncated infinite series is more accurate to start with the smallest terms.
// RiemannZetaFunctionWin32Console.c (c) Saturday, July 13-15, 2024
// James Pate Williams, Jr. some computations use 999,999,998 terms
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
typedef signed long long sll;
// https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function#cite_note-7
long double EvenZeta(int n)
{
sll A[12] = { 0ll,
6ll, 90ll, 945ll, 9450ll, 93555ll, 638512875ll,
18243225ll, 325641566250ll, 38979295480125ll,
1531329465290625ll, 13447856940643125ll };
sll B[12] = { 0ll, 1ll, 1ll, 1ll, 1ll, 1ll, 691ll, 2ll,
3617ll, 43867ll, 174611ll, 155366ll };
long double pi = 4.0 * atan(1.0);
return (long double)B[n / 2] * powl(pi, n) / (long double)A[n / 2];
}
long double Zeta(long s, long *terms)
{
int n = 0;
long double sum = 0.0;
*terms = 0;
for (n = 1000000000; n >= 2; n--)
{
sum += 1.0 / powl(n, s);
(*terms)++;
}
return sum + 1.0;
}
void PrintEvenZetaValue(int n)
{
clock_t time0 = clock();
long double ez = EvenZeta(n);
clock_t time1 = clock();
double runtime = ((double)time1 - time0) / CLOCKS_PER_SEC;
printf_s("Runtime in seconds to compute Zeta(%ld) = %16.15Lf: %Lf\n",
n, ez, runtime);
}
void PrintZetaValue(int n)
{
clock_t time0 = clock();
long terms = 0;
long double zeta = Zeta(n, &terms);
clock_t time1 = clock();
double runtime = ((double)time1 - time0) / CLOCKS_PER_SEC;
printf_s("Runtime in seconds to compute Zeta(%ld) = %16.15Lf terms = %ld: time = %Lf\n",
n, zeta, terms, runtime);
}
int main()
{
PrintEvenZetaValue(2);
PrintEvenZetaValue(4);
PrintEvenZetaValue(6);
PrintEvenZetaValue(8);
PrintZetaValue(2);
PrintZetaValue(4);
PrintZetaValue(6);
PrintZetaValue(8);
PrintZetaValue(3);
PrintZetaValue(5);
PrintZetaValue(7);
PrintZetaValue(9);
return 0;
}
Blog Entry Friday, July 12, 2024, © James Pate Williams, Jr.
I was performing an Internet search for prime counting functions, and I ran across the following PDF:
It was made available online in April of 2021. That made me remember the opening line of Geoffery Chaucer’s “Canterbury Tales”:
Whan that Aprill with his shoures soote
I seem to recall that in my English literature textbook ‘with’ was replaced with the German ‘mit’. Middle English was related to the old French and German languages I seem to recall.
Blog Entry Wednesday, July 10, 2024, © James Pate Williams, Jr. My Dual Interests in Cryptography and Number Theory
I became fascinated with secret key cryptography as a child. Later, as an adult, in around 1979, I started creating crude symmetric cryptographic algorithms. I became further enthralled with cryptography and number theory in 1996 upon reading Applied Cryptography, Second Edition: Protocols, Algorithms, and Source Code in C by Bruce Schneier and later the Handbook of Applied Cryptography by Alfred J. Menezes, Paul C. van Oorschot, and Scott A. Vanstone. After implementing many of the algorithms in both tomes, I communicated my results to two of the authors namely Bruce Schneier and Professor Alfred J. Menezes. In 1997 I developed a website devoted to constraint satisfaction problems and their solutions, cryptography, and number theory. I posted legal C and C++ source code. Professor Menezes advertised my website along with his treatise. See the following blurb:
In the spirit of my twin scientific infatuations, I offer yet another C integer factoring implementation utilizing the Free Large Integer Package (known more widely as lip) which was created by Arjen K. Lenstra (now a Professor Emeritus). This implementation includes Henri Cohen’s Trial Division algorithm, the Brent-Cohen-Pollard rho method, the Cohen-Pollard p – 1 stage 1 method, and the Lenstra lip Elliptic Curve Method. If I can get the proper authorization, I will later post the source code.
total time required for initialization: 0.056000 seconds
enter number below:
2^111+2
== Menu ==
1 Trial Division
2 Pollard-Brent-Cohen rho
3 p - 1 Pollard-Cohen
4 Lenstra's Elliptic Curve Method
5 Pollard-Lenstra rho
1
2596148429267413814265248164610050
number is composite
factors:
total time required factoring: 0.014000 seconds:
2
5 ^ 2
41
397
2113
enter number below:
0
total time required for initialization: 0.056000 seconds
enter number below:
2^111+2
== Menu ==
1 Trial Division
2 Pollard-Brent-Cohen rho
3 p - 1 Pollard-Cohen
4 Lenstra's Elliptic Curve Method
5 Pollard-Lenstra rho
2
2596148429267413814265248164610050
number is composite
factors:
total time required factoring: 1.531000 seconds:
2
5 ^ 2
41
397
2113
415878438361
3630105520141
enter number below:
0
total time required for initialization: 0.055000 seconds
enter number below:
2^111+2
== Menu ==
1 Trial Division
2 Pollard-Brent-Cohen rho
3 p - 1 Pollard-Cohen
4 Lenstra's Elliptic Curve Method
5 Pollard-Lenstra rho
3
2596148429267413814265248164610050
number is composite
factors:
total time required factoring: 0.066000 seconds:
2
5 ^ 2
41
838861
415878438361
3630105520141
enter number below:
0
total time required for initialization: 0.056000 seconds
enter number below:
2^111+2
== Menu ==
1 Trial Division
2 Pollard-Brent-Cohen rho
3 p - 1 Pollard-Cohen
4 Lenstra's Elliptic Curve Method
5 Pollard-Lenstra rho
4
2596148429267413814265248164610050
number is composite
factors:
total time required factoring: 0.013000 seconds:
2
5
205
838861
415878438361
3630105520141
enter number below:
0
Numerical Explorations 