要处理有理数而又不损失Matlab中的计算精度? [英] To handle rational number without losing accuracy of computation in Matlab?

问题描述

我想在计算中使用这个有理数而又不损失Matlab中图片的准确性:

I want to use this rational number in computations without losing the accuracy of the picture in Matlab:

f = 359.0 + 16241/16250.0

我认为例如f = uint64(359.0 + 16241/16250.0)的存储会失去准确性，在Matlab中被视为360.

I think storing, for instance by f = uint64(359.0 + 16241/16250.0) loses accuracy, seen as 360 in Matlab.

我认为处理事情的最好方法是永远不要存储价值，而要存储诸如此类的因素

I think the best way to handle the thing is never to store the value but to store its factors like

% f = a + b/c
a = 359
b = 16241
c = 16250

然后使用变量a，b和c进行计算，并将结果显示为图片.

and then doing computation by the variables a, b and c, and giving the result as a picture.

这是保持准确性的好方法吗?

推荐答案

如您所建议，如果您绝对不希望在存储有理数时失去准确性，那么最好的解决方案可能是按照数字的形式存储整数分量.

As you suggest, if you absolutely don't want to lose accuracy when storing a rational number, the best solution probably is to store the number in terms of its integer components.

除了将三个组成部分(f = a + b/c)之外，还可以将表示减少为两个组成部分:f = n/d.因此，每个有理数将被定义(并存储)为两个分量的整数矢量[n d].例如，示例中的数字f对应于n=5849991和d=16250.

Instead of your three components (f = a + b/c) you can reduce the reprentation to two components: f = n/d. Thus each rational number would be defined (and stored) as the two-component integer vector [n d]. For example, the number f in your example corresponds to n=5849991 and d=16250.

为简化处理以这种方式存储的有理数的方法，可以定义一个辅助函数，该函数在应用所需的操作之前将从[n d]表示形式转换为n/d:

To simplify handling rational numbers stored this way, you could define a helper function which converts from the [n d] representation to n/d before applyling the desired operation:

useInteger = @(x, nd, fun) fun(x,double(nd(1))/double(nd(2)));

然后

>> x = sqrt(pi);
>> nd = int64([5849991 16250]);
>> useInteger(x, nd, @plus)
ans =
  361.7719
>> useInteger(x, nd, @times)
ans =
  638.0824

如果要在计算中实现任意高精度，则应考虑使用变量字符串参数的高精度算术(vpa).使用这种方法，您可以指定所需的位数:

If you want to achieve arbitrarily high precision in computations, you should consider using variable-precision arithmetic (vpa) with string arguments. With that approach you get to specify how many digits you want:

>> vpa('sqrt(pi)*5849991/16250', 50)
ans =
638.08240465923757600307902117159072301901656248436

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