MatLab-可变精度算法 [英] MatLab - variable precision arithmetic

问题描述

我有一个简短的问题，关于vpa命令，该命令可用于评估MatLab中的符号表达式.

I have a brief question regarding the vpa command one may use to evaluate symbolic expressions in MatLab.

我的教科书上说:

在数字上使用诸如sqrt之类的功能时，请务必小心，默认情况下会产生双精度浮点数.您需要将此类输入作为符号字符串传递给vpa正确的评估:vpa('sqrt(5)/pi')."

"You need to be careful when you use functions such as sqrt on numbers, which by default result in a double-precision floating-point number. You need to pass such input to vpa as a symbolic string for correct evaluation: vpa('sqrt(5)/pi')."

我不太理解这里的行话.为什么对于大多数输入我键入vpa(input)或vpa('input')都得到完全相同的答案，而对平方根却没有?例如，如果我键入vpa(sin(pi/4))或vpa('sin(pi/4)')，我得到的答案完全相同，但是如果我将上面的给定问题键入为vpa(sqrt(5)/pi)，则得到的答案将与我键入vpa('sqrt(5)/pi')时得到的答案不同.

I don't quite understand the jargon here. Why is it that for most inputs I get the exact same answer whether I type vpa(input) or vpa('input'), but not for square roots? For instance, if I type vpa(sin(pi/4)) or vpa('sin(pi/4)'), I get the exact same answers, but if I type the give problem above as vpa(sqrt(5)/pi), I do not get the same answer as when I type vpa('sqrt(5)/pi').

如果有人能比我的书更详细地解释这一点，我将不胜感激！

If someone could explain this in a little more detail than what my book does above, I would be very grateful!

推荐答案

我不是MatLab专家，但是没有引号，您正在将sqrt(5)/pi的结果传递到vpa()中:

I'm no MatLab expert, but without quotes, you're passing the result of sqrt(5)/pi into vpa():

  vpa(sqrt(5)/pi)
= vpa(0.7117625434171772)

用引号引起来，您要将表达式 sqrt(5)/pi(未经评估并以精确形式)传递到vpa()中，然后告诉MatLab以可变的精度计算sqrt(5)/pi.

With quotes, you're passing the expression sqrt(5)/pi (unevaluated and in exact form) into vpa() and then telling MatLab to compute sqrt(5)/pi with variable precision.

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