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my wolfram version: 13.1

This is a very simple equation, and I want to solve for "A".

$$ \frac{A^2 \pi^{3 / 2}}{2 \sqrt{2} a^{3 / 2}}=1 $$

And I'm sure "A" is positive, so his solution is

$$ A=\left(\frac{2}{\pi}\right)^{3 / 4} a^{3 / 4} $$

As you can see when I solve this equation directly, I can't solve it

Clear["Global`*"];
$Assumptions = a > 0

Solve[(A^2Pi^(3/2))/(2Sqrt[2]*a^(3/2)) == 1, A]

enter image description here

To solve this problem, I replaced "A^2" with "A", and then I took the square root of "A" and got the answer

enter image description here

I'm confused about this

我心永恒
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    With v13.1 on a Mac I get {{A -> (2*Sqrt[2]*a^(3/2))/Pi^(3/2)}}. If the domain is specified as Reals then {{A -> ConditionalExpression[ (2*Sqrt[2]*a^(3/2))/Pi^(3/2), a > 0]}} – Bob Hanlon Mar 30 '23 at 15:24
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    Did you mean to have A or A^2 in your first code snippet? It's A^2 in the screenshot but A in the code. – Michael Seifert Mar 30 '23 at 20:58
  • @MichaelSeifert In the first equation, I used "A^2", I tried to find A, got A positive number, failed, so I replaced "A^2" with "a", and took the square root – 我心永恒 Mar 31 '23 at 00:46
  • Please add the code text for all of the code sample to your question. Currently the question looks confusing (at first glance, at least). And user64494 is clearly fooled. – xzczd Mar 31 '23 at 00:55
  • 以防你没听懂，我现在用汉语重复一遍：你贴错代码了。 – xzczd Mar 31 '23 at 01:43
  • @xzczd I have do it – 我心永恒 Mar 31 '23 at 06:23
  • Omit $Assumptions = a > 0, and the desired solution results. – bbgodfrey Mar 31 '23 at 12:07
  • I am posting this comment in two forms: English and "Google "translated mainland Chinese. It has been very unclear whether the "A" version or the "A^2"version is the actual problem with which you desired assistance. To receive answer in form that you seem to want, you have tell Mathematica what you desire: ToRadicals[Solve[(APi^(3/2))/(2Sqrt[2]a^(3/2)) == 1 && A >= 0, A, Reals, MaxExtraConditions -> All]] which produces in Mathematica 13.2.1 the that answer: {{A -> ConditionalExpression[(2Sqrt[2]*a^(3/2))/Pi^(3/2), a > 0]}}. –  Mar 31 '23 at 15:08
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    @我心永恒 我以两种形式发布此评论：英文和“Google”翻译的大陆中文。 目前还不清楚“A”版本或“A^2”版本是否是您需要帮助的实际问题。 要以您似乎想要的形式接收答案，您已经告诉 Mathematica 您想要什么： ToRadicals[Solve[(APi^(3/2))/(2Sqrt[2]a^(3/2) ) == 1 && A >= 0, A, Reals, MaxExtraConditions -> All]] 在{{A -> ConditionalExpression[(2Sqrt[2]*a^(3/2))/Pi^(3/2), a > 0]}}。 –  Mar 31 '23 at 15:16
  • @我心永恒 I suggest reading how to enter usable code from notebook and common pitfalls and misunderstanding of Wolfram Mathematica. Also, remember that Mathematica generally works in complex mode. The documentation almost always when reals are assumed. –  Mar 31 '23 at 15:54
  • @我心永恒 我建议阅读[如何从笔记本输入可用代码](https://mathematica.meta.stackexchange.com/questions/1584/how-to-copy-code-from-mathematica-so-it-looks-good-on -this-site) 和 Wolfram Mathematica 的常见陷阱和误解。 另外，请记住 Mathematica 通常在复数模式下工作。 假设实数时，文档几乎总是如此。 –  Mar 31 '23 at 15:55

3 Answers3

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This behavior seems to exist for version 12.2 and later. You don't mention any assumptions you are making, but this is what I get with different assumptions in version 13.2.1.

$Assumptions = a > 0

Solve[(A^2Pi^(3/2))/(2Sqrt[2]*a^(3/2)) == 1, A]

The result matches your result, but with

$Assumptions =.

Solve[(A^2Pi^(3/2))/(2Sqrt[2]a^(3/2)) == 1, A]
({{A -> (2/Pi)^(3/4)(-a^(3/4))}, {A -> (2/Pi)^(3/4)a^(3/4)}}*)

The result is what you would expect. Prior to version 12.2, the result is what you would expect in both cases. This new behavior has broken a lot of my calculations.

Bill Watts
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Adding domain Reals,

Solve[(A \[Pi]^(3/2))/(2 Sqrt[2] a^(3/2)) == 1, A, Reals]

{{A -> ConditionalExpression[(2 Sqrt[2] a^(3/2))/\[Pi]^(3/2), a > 0]}}

Addition. In 13.2 on Windows 10

Solve[(A \[Pi]^(3/2))/(2 Sqrt[2] a^(3/2)) == 1, A]

{{A -> (2 Sqrt[2] a^(3/2))/\[Pi]^(3/2)}}

The above answer is generic. Also Solve may do non-equivalent transformations of the original equation.

user64494
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It works for me in Wolfram Cloud (just a few minutes ago). I am on my cellphone which why I used Wolfram Cloud. In my computer I use 13.2.1.

{{A -> -a^(3/4)(2/π)^(3/4)}, {A -> a^(3/4)(2/π)^(3/4)}}

What version of Mathematica are you using?

By the way, your output is a solution. I suggest using ToRadicals to see the solution in the form you expected.

  • Are you sure? What is the code used by you? – user64494 Mar 30 '23 at 07:39
  • my wolfram version is 13.1 – 我心永恒 Mar 30 '23 at 07:47
  • @我心永恒 Yes, I executed the Solve[A^2 ... version in Wolfram Cloud. That is the answer it returned. My computer is shut down for the night. –  Mar 30 '23 at 07:55
  • @IAmANaïf If I want to get {A -> a^(3/4)(2/π)^(3/4)},but do not want to use [[2]]，what should I do? – 我心永恒 Mar 30 '23 at 07:56
  • @user64494 Quite sure. To get the first answer that 我心永恒 received the original Solve command had to have started Solve[A^2 ... and that is the version that returned the answer I gave. –  Mar 30 '23 at 08:00
  • @我心永恒 ToRadials[Solve[(A^2 [Pi]^(3/2))/(2 Sqrt[2] a^(3/2)) == 1, A]] –  Mar 30 '23 at 08:03
  • @IAmANaïf will get {{A -> I a^(3/4) (2/\[Pi])^(3/4)}, {A -> -a^(3/4) (2/\[Pi])^(3/4)}} – 我心永恒 Mar 30 '23 at 08:08
  • i just redid Solve[(A^2 [Pi]^(3/2))/(2 Sqrt[2] a^(3/2)) == 1, A], Wolfram Cloud returned {{A->-(a^(3and WolframAlpha returned A = (4 sqrt(2) a^(3/2))/π^(3/2) and sqrt(a)!=0. I copied the Wolfram Cloud answer by hand and erroneously. Mea Culpa. Changing Solve –  Mar 30 '23 at 08:50
  • I just redid Solve[(A^2 [Pi]^(3/2))/(2 Sqrt[2] a^(3/2)) == 1, A], Wolfram Cloud returned {{A->-(a^(3/4)(2/Pi)^(3/4))},{A->a^(3/4)(2/Pi)^(3/4)}} and WolframAlpha returned A = (4 sqrt(2) a^(3/2))/π^(3/2) and sqrt(a)!=0. I copied the Wolfram Cloud answer by hand and erroneously. Mea Culpa. Changing Solve to Reduce in Wolfram Cloud returned (Sqrt[a]!=0&&A==-(a^(3/4)(2/Pi)^(3/4)))||(Sqrt[a]!=0&&A==a^(3/4)(2/Pi)^(3/4)). Solve in Wolfram Cloud failed to notice that $a$ was originally in the denominator and therefore must not be $0$ (division by $0$ error). –  Mar 30 '23 at 09:19
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    …Why do you post (almost) the same answer twice?: https://mathematica.stackexchange.com/a/283019/1871 – xzczd Mar 30 '23 at 09:49
  • @xzczd I regret that I had a cellphone crash in the midst of a comment entry. Last night's work was all on my cellphone. Do not be concerned overly. I was annoyed as well. I had two users both commenting at me at cross purposes. Lastly, there seemed to be problems communicating. –  Mar 30 '23 at 22:13