[读者文库]Java Number Cruncher

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楼主

ReneeBK 发表于 2015-6-6 09:23:38 |AI写论文

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Java Number Cruncher: The Java Programmer's Guide to Numerical Computing
By: Ronald Mak
Publisher: Prentice Hall
Pub. Date: October 29, 2002
Print ISBN-10: 0-13-046041-9
Print ISBN-13: 978-0-13-046041-7
Pages in Print Edition: 480
Subscriber Rating: [0 Ratings]

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关键词：crunch Number Java CRU UNC Java 读者

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沙发

ReneeBK 发表于 2015-6-6 09:24:53

package numbercruncher.program2_2;
/**
* PROGRAM 2-2: Integer Wrapper Classes
*
* Print the values of some of the constants
* defined in the integer wrapper classes.
*/
public class IntegerWrapperClasses
{
public static void main(String args[])
{
// Byte
System.out.println("Byte.MIN_VALUE = " + Byte.MIN_VALUE);
System.out.println("Byte.MAX_VALUE = " + Byte.MAX_VALUE);
System.out.println();
// Short
System.out.println("Short.MIN_VALUE = " + Short.MIN_VALUE);
System.out.println("Short.MAX_VALUE = " + Short.MAX_VALUE);
System.out.println();
// Character
System.out.println("Character.MIN_VALUE = " +
(int) Character.MIN_VALUE);

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Output:
Byte.MIN_VALUE = -128
Byte.MAX_VALUE = 127
Short.MIN_VALUE = -32768
Short.MAX_VALUE = 32767
Character.MIN_VALUE = 0
Character.MAX_VALUE = 65535
Integer.MIN_VALUE = -2147483648
Binary: 10000000000000000000000000000000
Hex: 80000000
Integer.MAX_VALUE = 2147483647
Binary: 1111111111111111111111111111111
Hex: 7fffffff
Long.MIN_VALUE = -9223372036854775808
Binary: 1000000000000000000000000000000000000000000000000000000000000000
Hex: 8000000000000000
Long.MAX_VALUE = 9223372036854775807
Binary: 111111111111111111111111111111111111111111111111111111111111111
Hex: 7fffffffffffffff

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藤椅

ReneeBK 发表于 2015-6-6 09:25:51

Listing 2-3a. Class IntPower for doing integer exponentiation.

Listing 2-3a. Class IntPower for doing integer exponentiation.
Code View: Scroll / Show All
package numbercruncher.mathutils;
/**
* Raise a double value to an integer power.
*/
public final class IntPower
{
/**
* Compute and return x^power.
* @param x
* @param power the integer power
* @return x^power
*/
public static double raise(double x, int exponent)
{
if (exponent < 0) return 1/raise(x, -exponent);
double power = 1;
// Loop to compute x^exponent.
while (exponent > 0) {
// Is the rightmost exponent bit a 1?
if ((exponent & 1) == 1) power *= x;

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板凳

ReneeBK 发表于 2015-6-6 09:27:03

Listing 2-3b. A test program for class IntPower.
Code View: Scroll / Show All
package numbercruncher.program2_3;
import numbercruncher.mathutils.IntPower;
/**
* PROGRAM 2-3: Test Class IntPower
*
* Test the IntPower class.
*/
public class TestIntPower
{
public static .void main(String args[])
{
System.out.println(IntPower.raise(2, 5) + " " +
Math.pow(2, 5));
System.out.println(IntPower.raise(2, -5) + " " +
Math.pow(2, -5));
System.out.println(IntPower.raise(2, 0) + " " +
Math.pow(2, 0));
System.out.println(IntPower.raise(2.5, 5) + " " +
Math.pow(2.5, 5));
System.out.println();
System.out.println(IntPower.raise(-2, 5) + " " +
Math.pow(-2, 5));
System.out.println(IntPower.raise(-2, -5) + " " +
Math.pow(-2, -5));
System.out.println(IntPower.raise(-2, 0) + " " +
Math.pow(-2, 0));
System.out.println(IntPower.raise(-2.5, 5) + " " +
Math.pow(-2.5, 5));
}
}

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Output:
32.0 32.0
0.03125 0.03125
1.0 1.0
97.65625 97.65625
-32.0 -32.0
-0.03125 -0.03125
1.0 1.0
-97.65625 -97.65625

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报纸

ReneeBK 发表于 2015-6-6 09:28:16

Listing 4-5a. Kahan's Summation Algorithm for the float type.
Code View: Scroll / Show All
package numbercruncher.mathutils;
/**
* Implement Kahan's Summation Algorithm for the float type.
*/
public class KahanSummation
{
/** the current running sum */ private float sum;
/** the current correction */ private float correction;
/** the current corrected addend */ private float correctedAddend;
/**
* Constructor.
*/
public KahanSummation() {}
/**
* Return the current corrected value of the running sum.
* @return the running sum's value
*/
public float value() { return sum + correction; }
/**
* Return the corrected value of the current addend.
* @return the corrected addend value

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地板

ReneeBK 发表于 2015-6-6 09:29:04

Listing 4-5b. Fraction sums for n = 100,000,000 by the Kahan Summation Algorithm.
Code View: Scroll / Show All
package numbercruncher.program4_5;
import numbercruncher.mathutils.KahanSummation;
import numbercruncher.mathutils.AlignRight;
/**
* PROGRAM 4-5: Fraction Sum 100M by the Kahan Summation Algorithm
*
* Use the Kahan Summation Algorithm to compute
* the sum 1/d + 2/d + 3/d + ... + n/d = d/d
* where:
* n = 100,000,000
* d = 1 + 2 + 3 + ... + n = (n/2)(n + 1)
*
* See if the sum is closer to 1.
*/
public class FractionSum100MKahan
{
private static final int GROUPS = 20;
private static final int MAX = 100000000; // 100M
public static void main(String args[])
{
AlignRight ar = new AlignRight();

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Output:
Code View: Scroll / Show All
i Running sum % ExpDiff>24
-----------------------------------------
5000000 0.0025000004 0
10000000 0.010000001 0
15000000 0.0225 0
20000000 0.04 0
25000000 0.0625 0
30000000 0.09 26
35000000 0.1225 12
40000000 0.16 46
45000000 0.2025 42
50000000 0.25 36
55000000 0.3025 64
60000000 0.35999998 61
65000000 0.42249998 58
70000000 0.48999998 54
75000000 0.5625 72
80000000 0.64 73
85000000 0.72249997 72
90000000 0.81 70
95000000 0.9025 68
100000000 1.0 67
% error = 0.0

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7楼

ReneeBK 发表于 2015-6-6 09:30:13

Listing 4-9. Computation of e−19.5 using the Taylor series.
Code View: Scroll / Show All
package numbercruncher.program4_9;
import numbercruncher.mathutils.AlignRight;
/**
* PROGRAM 4-9: e to x
*
* Compute e^x using the Taylor series with x = -19.5
* The final value should be approximately 3.4e-9
*/
public class EtoX
{
private static final double x = -19.5;
public static void main(String args[])
{
AlignRight ar = new AlignRight();
int k = 0;
double numerator = 1;
double denominator = 1;
double sum = 1; // running sum
double prevSum = 0; // previous value of running sum
ar.print("k", 2);

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Output:
Code View: Scroll / Show All
k Numerator
Denominator Fraction Running sum
-------------------------------------------------------------------------
1 -19.5
1.0 -19.5 -18.5
2 380.25
2.0 190.125 171.625
3 -7414.875
6.0 -1235.8125 -1064.1875
4 144590.0625
24.0 6024.5859375 4960.3984375
...
17 -8.522919630788428E21
3.55687428096E14 -2.3961824224183973E7 -1.2635486167179534E7
18 1.6619693280037435E23
6.402373705728E15 2.595864290953264E7 1.3323156742353106E7
19 -3.2408401896073E24
1.21645100408832E17 -2.6641765091362443E7 -1.3318608349009337E7
20 6.319638369734235E25
2.43290200817664E18 2.5975720964078385E7 1.2657112615069048E7
...
49 -1.6281557854172669E63
6.082818640342675E62 -2.676646932425302 -0.754026189305919
50 3.1749037815636704E64
3.0414093201713376E64 1.0438923036458678 0.2898661143399488

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8楼

ReneeBK 发表于 2015-6-6 09:31:21

Listing 4-10. Computation of e−19.5 by inverting the result of the Taylor series at x = +19.5.
Code View: Scroll / Show All
package numbercruncher.program4_10;
import numbercruncher.mathutils.AlignRight;
/**
* PROGRAM 4-10: e to x Inverse
*
* Compute e^x at x = -19.5 by using the Taylor series
* with x = 19.5 and then taking the inverse of the result.
* The final value should be approximately 3.4e-9
*/
public class EtoXInverse
{
private static final double x = -19.5;
public static void main(String args[])
{
AlignRight ar = new AlignRight();
int k = 0;
double numerator = 1;
double denominator = 1;
double sum = 1; // running sum
double prevSum = 0; // previous value of running sum
double xInverse = -x;

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Output:
Code View: Scroll / Show All
k Numerator
Denominator Fraction Running sum
-------------------------------------------------------------------------
1 19.5
1.0 19.5 20.5
2 380.25
2.0 190.125 210.625
3 7414.875
6.0 1235.8125 1446.4375
4 144590.0625
24.0 6024.5859375 7471.0234375
...
63 1.871459856776355E81
1.98260831540444E87 9.4393826669419E-7 2.9426756604150844E8
64 3.649346720713892E82
1.2688693218588417E89 2.87606190633386E-7 2.9426756604150873E8
65 7.116226105392089E83
8.247650592082472E90 8.628185719001579E-8 2.942675660415088E8
66 1.3876640905514574E85
5.443449390774431E92 2.5492366897050122E-8 2.942675660415088E8
e^-19.5 = 3.3982678194950715E-9
% error = 1.217062127663519E-14

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9楼

ReneeBK 发表于 2015-6-6 09:31:56

Listing 4-11. Computation of e−19.5 by splitting the exponent before using the Taylor series.
Code View: Scroll / Show All
package numbercruncher.program4_11;
import numbercruncher.mathutils.IntPower;
import numbercruncher.mathutils.AlignRight;
/**
* PROGRAM 4-11: e to x with Split Exponent
*
* Compute e^x by splitting the exponent x into its whole and
* fraction components before using the Taylor series.
*/
public class EtoXSplit
{
private static final double x = -19.5;
private static AlignRight ar = new AlignRight();
public static void main(String args[])
{
double result;
// Split off the whole part of |x|.
double xAbs = Math.abs(x);
int xWhole = (int) xAbs;

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10楼

ReneeBK 发表于 2015-6-6 09:33:16

Listing 5-0d. The abstract base class for the subclasses that implement root-finding algorithms.
Code View: Scroll / Show All
package numbercruncher.mathutils;
/**
* Abstract base class for the root finder classes.
*/
public abstract class RootFinder
{
/** the function whose roots to find */ protected Function function;
/** iteration counter */ private int n;
/** maximum number of iterations */ private int maxIters;
/**
* Constructor.
* @param function the function whose roots to find
* @param maxIters the maximum number of iterations
*/
public RootFinder(Function function, int maxIters)
{
this.function = function;
this.maxIters = maxIters;
}
/**
* Check the interval.
* @param xMin x-coordinate of the left of the interval

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