Matlab exponent bug?
3 ビュー (過去 30 日間)
古いコメントを表示
Hello MatLabers
I have encountered a problem when calculating a non integer exponent/power of a variable.
example:
>> -3.^(1.3)
>> ans = -4.1712
thats exactly what I aim to calculate. However, if I do the exact same thing with a variable - the result gets complex:
>> a = -3
>> a.^(1.3)
>> ans = -2.4518 - 3.3745i
is this an known issue or am I doing somehting wrong? (tested on R2010a, R2011b)
採用された回答
Daniel Shub
2011 年 11 月 24 日
This is a precedence issue: http://www.mathworks.com/help/releases/R2011a/techdoc/matlab_prog/f0-40063.html#f0-38155
Try
(-3).^(1.3)
3 件のコメント
その他の回答 (2 件)
Jan
2011 年 11 月 24 日
The POWER operation has a higher precedence than the unary minus. Try this:
-3 .^ (1.3)
(-3) .^ (1.3)
Edgar An
2021 年 3 月 7 日
編集済み: Edgar An
2021 年 3 月 7 日
The problem is not about precedence. The problem occurs when we use variables instead of numbers. Just like the person who posted.
-0.685^1.5 gives a correct answer
but a = -0.685, b = 1.5, and then a^b gives wrong answer
i am using R2020b version
3 件のコメント
Walter Roberson
2021 年 3 月 7 日
In MATLAB, a^b is defined to be equivalent to exp(log(a)*b). When a is negative the log is complex with a πι component and if b is not an integer then the exp() of the πι*b is going to be complex.
In practice you can tell from timings that for at least some integer values a^b is not implemented through logs: for example a^2 has timing the same as a*a, but the principle is the same.
Bruno Luong
2021 年 3 月 7 日
編集済み: Bruno Luong
2021 年 3 月 7 日
To be precise for z complex (including negarive real)
log(z)
is defined in term of real-argument function log and atan2
log(abs(z)) + 1i*angle(z)
where abs(z) is
sqrt(imag(z)^2+real(z)^2) % the square "x^2" here is interpreted as (x*x)
angle(z) is
atan2(imag(z),real(z))
with all the rule we discuss recently, notably discontinuity when real(z) is negative and numeriral sign of imag(z).
The definition of
exp(z)
has no ambiguity.
参考
カテゴリ
Help Center および File Exchange で Logical についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!