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

Answer 1

[cheng sy]

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

[Torsten]

cosh(A) is evaluated elementwise.

If you want matrix exponential, use Y=(expm(S)+expm(-S))/2.

Best wishes

Torsten.

Answer 2

[cheng sy]

In factor, cosh is can be expended by the Maclaurin’s series:

Answer 3

[Walter Roberson]

"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.

Answer 4

[cheng sy]

thanks!