is the cosh(Hyperbolic cosine) wrong?

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In recent days, I have confused by the function of cosh(Hyperbolic cosine) in matlab. Suppose the matrix is the following:
S=[ 1.0e-05 *
-0.1293 + 0.0195i -0.0128 + 0.0079i -0.0144 + 0.0090i
-0.0141 + 0.0085i -0.1266 + 0.0197i -0.0141 + 0.0085i
-0.0144 + 0.0090i -0.0128 + 0.0079i -0.1293 + 0.0195i]
The cosh of S with matlab function cosh, the result is:
ans =
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
In factor, cosh is can be expended by the Maclaurin’s series:
When x is a matrix, the first term of the Maclaurin’s series is unite matrix or identity matrix.
So the result should be :
ans =
1.0000 - 0.0000i 0.0000 - 0.0000i 0.0000 - 0.0000i
0.0000 - 0.0000i 1.0000 - 0.0000i 0.0000 - 0.0000i
0.0000 - 0.0000i 0.0000 - 0.0000i 1.0000 - 0.0000i

采纳的回答

cheng sy
cheng sy 2015-6-24
The cosh of S with matlab function cosh, the result is:
ans =
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
1.0000 - 0.0000i 1.0000 - 0.0000i 1.0000 - 0.0000i
  1 个评论
Torsten
Torsten 2015-6-24
cosh(A) is evaluated elementwise.
If you want matrix exponential, use Y=(expm(S)+expm(-S))/2.
Best wishes
Torsten.

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更多回答(3 个)

cheng sy
cheng sy 2015-6-24
In factor, cosh is can be expended by the Maclaurin’s series:

Walter Roberson
Walter Roberson 2015-6-24
"cosh(X) is the hyperbolic cosine of the elements of X."
In other words, cosh() is applied one by one to the elements of X, independently of the others.

cheng sy
cheng sy 2015-6-24
thanks!

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