is the cosh(Hyperbolic cosine) wrong?

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

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

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:
"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.

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