perl - perl 中的牛顿法

Question

好吧，在我的数学课上，我们被要求编写一个程序来执行和打印牛顿法，直到值收敛并且我们有函数的根。起初我以为这很容易。直到我无法获得第二次使用的第一次使用的值。我的语言知识是基本的。非常基本，所以您将要看到的可能并不漂亮。

#!usr/bin/perl

use PDL;

print "First guess? (this is x0)\n";
$xorig = <>;

do {
  &fx;
} until ($fex == 0);

sub fx {

  if ($xn == 0) {
    $x = $xorig;
  }
  else {
    $x = $xn;
  }

  print "What is the coefficient (for each factor) of your function?\n";
  $fcx = <STDIN>;
  push @coefficient_of_x, $fcx;

  print "... times x to the (enter exponent, if no exponent, enter 1. if no x, enter 0)?\n";
  $fex = <STDIN>;
  push @exponent_x, $fex;

  chomp ($fcx, $fex, $x, $xorig);

  $factor = $fcx * ($x ** $fex);
  push @fx, $factor;
}

my $fx = 0;
foreach my $variable (@fx) {
  $fx = $variable + $fx #THIS PROVIDES A VALUE FOR THE GIVEN F(X) WITH A GIVEN X VALUE
}
print "f($x)=$fx\n";

do {
  &fprimex;
} until ($fprimeex == 0);

sub fprimex {

  if ($xn == 0) {
    $x = $xorig;
  }
  else {
    $x = $xn;
  }

  print "What is the coefficient (for each factor) of your derivative function?\n";
  $fprimecx = <STDIN>;
  push @coefficient_of_fpx, $fprimecx; 

  print "... times x to the (enter exponent, if no exponent, enter 1. if no x, enter 0)?\n";
  $fprimeex = <STDIN>;
  push @exponent_fpx, $fprimeex;

  chomp ($fprimecx, $fprimeex, $x, $xorig);

  $factorprime = $fprimecx * ($x ** $fprimeex);
  push @fprimex, $factorprime;
}

$fprimex = 0;
foreach my $variableprime (@fprimex) {
  $fprimex = $variableprime + $fprimex #THIS PROVIDES A VALUE FOR THE GIVEN F'(X) WITH THAT SAME X VALUE
}
print "f'($x)=$fprimex\n";

sub x0 {
  $xn = $xorig - $fx / $fprimex; #THIS IS NEWTON'S METHOD EQUATION FOR THE FIRST TIME
  push @newxn, $xn;
  print "xn ia $xn\n";
}

&x0;

foreach $value (@exponent_x) {
  $exponent_x = $xn ** $value;
  push @part1, $exponent_x;
  $part1 = @part1;
}

foreach $value2 (@coefficient_of_x) {
  $part2 = $value2 * @part1;
  push @final1, $part2;
}

print "@part1\n";
print "@final1\n";

基本上是我先问第一个猜测。我使用这个值来定义 f(x) 的系数和指数，以根据给定的 x 获得 f(x) 的值。我为 f'(x) 再做一次。然后我第一次执行牛顿法得到新值xn。但是我很难获得 f(xn) 和 f'(xn) 的值，这意味着我无法获得 x(n+1) 并且无法继续牛顿的方法。我需要帮助。

Answer 1

欢迎使用 Perl。

我强烈建议对您的代码进行以下更改:

1. 始终包含 use strict;和 use warnings;在每个 Perl 脚本中。

2. 始终 chomp您从 STDIN 输入的内容作为您的输入:

   chomp( my $input = <STDIN> );
   

3. 不要不必要地创建子例程，尤其是对于像这样的一次性脚本。

4. 而不是使用 statement modifier do 的形式，我建议使用无限 while用循环控制语句退出:

   while (1) {
   
       last if COND;
   }
   

5. 最后，由于多项式的系数都与 X 的指数相关联，我建议使用 %hash 来方便地保存这些值。

如图所示:

#!usr/bin/env perl
use strict;
use warnings;

print "Build your Polynomial:\n";

my %coefficients;

# Request each Coefficient and Exponent of the Polynomial
while (1) {
    print "What is the coefficient (for each factor) of your function? (use a bare return when done)\n";
    chomp( my $coef = <STDIN> );

    last if $coef eq '';

    print "... times x to the (enter exponent, if no exponent, enter 1. if no x, enter 0)?\n";
    chomp( my $exp = <STDIN> );

    $coefficients{$exp} = $coef;
}

print "\nFirst guess? (this is x0)\n";
chomp( my $x = <> );

# Newton's Method Iteration
while (1) {
    my $fx  = 0;
    my $fpx = 0;

    while ( my ( $exp, $coef ) = each %coefficients ) {
        $fx += $coef * $x**$exp;
        $fpx += $coef * $exp * $x**( $exp - 1 ) if $exp != 0;
    }

    print "   f(x) = $fx\n";
    print "   f'(x) = $fpx\n";

    die "Slope of 0 found at $x\n" if $fpx == 0;

    my $new_x = $x - $fx / $fpx;

    print "Newton's Method gives new value for x at $new_x\n";

    if ( abs($x - $new_x) < .0001 ) {
        print "Accuracy reached\n";
        last;
    }

    $x = $new_x;
}