Why the answer is like that?

2018 3 2
3 个回答
回答已采纳
更新时间：2018 3 3
6 次查看（30 天）

0 个投票

syms x
s = vpa(solve((0.5)==(log(5)*log10(8)*sinh(x))/(atan(0.5)*(3^(1/5))*abs(cos(2*pi/3))*exp(cos(2*pi/3))),x))
the correct result of this equation is 0,06 but when i run this command matlab gives me this
s =
0.060219526083180976165227748603359
- 0.060219526083180976165227748603359 + 3.1415926535897932384626433832795i
So, why i found this result and how can i found just " 0,06 " ?
Thank you

请先登录，再回答此问题。

 采纳的回答

Walter Roberson
Walter Roberson 2018-3-2

1 个投票

0.06 is not a solution to that equation. If you substitute 0.06 for x you get
0.5 == 0.49817509413836472619769716883263
which is false.
The exact solution is asinh((3^(1/5)*atan(1/2)*exp(-1/2)*log(10))/(4*log(5)*log(8)))

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Jan
Jan 2018-3-2
编辑：Jan 2018-3-2
@Walter: I opened this thread 20 minutes ago. To this time I did not see your answer, also you have posted it 16 minutes earlier. Do you observe such delays also?
It looks awkward that I have repeated your answer, but I did not see it.

请先登录，再进行评论。

更多回答（2 个）

Jan
Jan 2018-3-2

1 个投票

No, 0.06 is not a solution. Simply try it:
x = 0.06
(log(5)*log10(8)*sinh(x)) / (atan(0.5)*(3^(1/5))*abs(cos(2*pi/3))*exp(cos(2*pi/3)))
This is 0.498175094138365, and not 0.5 .
But the shown results match the equation. So please explain, why you assume that 0.06 is a solution. This assumption is the problem to be solved.
Sam Oznc
Sam Oznc 2018-3-2

0 个投票

First thank you for your answers i am really pleased.
This is tutorial question and yes i know, real answer is not 0.06 but question want to the answer with only 2 digit form(i think "format bank" command doing this), so it is 0.06 but i didnt ask that.
My question is why i found this answer and how i can find the correct answer with only two digit (like 0.xx) ?
s =
0.060219526083180976165227748603359
- 0.060219526083180976165227748603359 + 3.1415926535897932384626433832795i

类别

在 帮助中心 和 File Exchange 中查找有关 Get Started with MuPAD 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by