Why the answer is like that?

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Sam Oznc
Sam Oznc 2018 年 3 月 2 日
コメント済み: Sam Oznc 2018 年 3 月 3 日
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

採用された回答

Walter Roberson
Walter Roberson 2018 年 3 月 2 日
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 件のコメント
Jan
Jan 2018 年 3 月 2 日
編集済み: Jan 2018 年 3 月 2 日
@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.

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その他の回答 (2 件)

Jan
Jan 2018 年 3 月 2 日
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 年 3 月 2 日
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
  2 件のコメント
Walter Roberson
Walter Roberson 2018 年 3 月 2 日
vpa(s,2)
Sam Oznc
Sam Oznc 2018 年 3 月 3 日
That's cool. Thank you

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