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I am trying to check whether the limit mentioned in the equation (3.5) of this textbook (page 14 of the PDF) does really simplify into the claimed expression.

I am trying to simplify this limit in particular: $$ \lim_{k\to\infty} \left(\mathbf{1} + \alpha_a\frac{X_a}{k}\right)^k = \sum_{m=0}^{\infty} \frac{1}{m!}\left(\alpha_aX_a\right)^m \equiv e^{\alpha_aX_a}, $$ where $X_a$ is a matrix and $\alpha_a$ is a scalar.

To make this case as simple as possible, I tried to simplify the limit using the following concrete matrix $X = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$.

The result of the limit simplification should then be $\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}$, as is mentioned in equation (3.6) of the textbook.

However, Instead of that result, MMA simplified the limit into the $2\times2$ identity matrix. This is the expression I entered into MMA:

A = {
  {0, -a},
  {a, 0}
}

Limit[(IdentityMatrix[2] + A/N)^N, N -> Infinity] // MatrixForm

How can I replicate the textbook results?

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    $\begingroup$ This may not matter, but avoid uppercase symbols in general, and particularly uppercase single letter symbols like A, N. N is a function in its own right (see its documentation. $\endgroup$
    – MarcoB
    Feb 4, 2022 at 2:58
  • $\begingroup$ Thanks for the heads up, I didn't know about this. Changing N to k and A to r did not change the result, as you wrote (i.e. MMA still simplified the expression to $\mathbf{I}$). $\endgroup$
    – jordi
    Feb 4, 2022 at 3:35

1 Answer 1

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Clear["Global`*"]

Use MatrixPower and MatrixFunction

A = {{0, -a}, {a, 0}};

Limit[MatrixPower[IdentityMatrix[2] + A/n, n], n -> Infinity] // 
  FullSimplify

(* {{Cos[a], -Sin[a]}, {Sin[a], Cos[a]}} *)

MatrixFunction[Exp, A]

(* {{Cos[a], -Sin[a]}, {Sin[a], Cos[a]}} *)
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    $\begingroup$ There is also MatrixExp[A] $\endgroup$
    – Carl Woll
    Feb 4, 2022 at 4:15
  • $\begingroup$ Thank you, using MatrixPower did the trick. I did not expect Mathematica to require using a special function for matrix powers (I thought the ^ operator was enough). How can I be sure in the future that MMA won't provide misleading results (like the identity matrix) when simplifying arbitrary expressions? Are there any other caveats with symbolic simplification that I should be aware of? $\endgroup$
    – jordi
    Feb 4, 2022 at 15:10
  • $\begingroup$ Power[matrix, n] and MatrixPower[matrix, n] are two completely different mathematical operations. Likewise f[matrix] and MatrixFunction[f, matrix] are radically different. Mathematica cannot guess which operation you want; you need to specify the intended operation. $\endgroup$
    – Bob Hanlon
    Feb 4, 2022 at 15:36
  • $\begingroup$ @BobHanlon Oh, I didn't realize there was a difference, but after reading your comment, it makes complete sense to me! So Power[] performs element-wise power, whereas MatrixPower[] performs repeated matrix multiplication. $\endgroup$
    – jordi
    Feb 4, 2022 at 20:39

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