Below is a question and its corresponding solution involving exponentiating a matrix, which I can't understand.
A boost can be written in the form $${x^\mu}^\prime=\Lambda_\nu^{{\mu}^\prime}x^\nu$$ Here $\Lambda_\nu^{{\mu}^\prime}$ can be viewed as the elements of a $4 \times 4$ matrix $\Lambda$. Show that for the standard boost in the $x$ direction $\Lambda_\nu^{{\mu}^\prime} = \exp(\psi K)$ where $K$ is a $4 \times 4$ matrix and $\tanh \psi = v/c$. What are the corresponding matrices for boosts in the $y$ and $z$ directions?
Here is the solution word for word:
The matrix, $\Lambda$ corresponding to the boost is $$\Lambda=\begin{pmatrix}\cosh\psi & -\sinh\psi & 0 &0 \\-\sinh\psi & \cosh\psi & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{pmatrix}=\exp(\psi K)$$ where $K$ is $$\begin{pmatrix}0 & -1 & 0 &0 \\-1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{pmatrix}$$ To see this use $K^{2n}=J=\mathrm{diag}(1,1,0,0)$ and $K^{2n+1}=K$ so that $$\begin{align}\exp(\psi K)&= I +\psi K +\frac{\psi^2 K^2}{2!}+\frac{\psi^3 K^3}{3!}+\cdots\tag{1}\\&=\cosh{\psi J}+\sinh{\psi J}+\mathrm{diag}(0,0,1,1)\tag{2}=\Lambda\end{align}$$ For boosts in the $y$ and $z$ direction $\Lambda=\exp(\psi K_y)$ and $\Lambda=\exp(\psi K_z),$ respectively with $$K_y=\begin{pmatrix}0 & 0 & -1 &0 \\0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{pmatrix}\qquad K_z=\begin{pmatrix}0 & 0 & 0 & -1 \\0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\-1 & 0 & 0 & 0\end{pmatrix}$$
The part I can't understand is how eqn $(2)$ follows from eqn $(1)$. I know that the identity $e^x=\cosh x +\sinh x$ is being used, and that the power series for $\cosh x$ and $\sinh x$ are being used. The problem is that eqn $(2)$ seems correct only for the even power series, $K^{2n}$. But for the odd powers, say $$K^3=K\ne J$$ so by that logic I think eqn $(2)$ should be $$\cosh{\psi J}+\sinh{\psi \color{red}{K}}+\mathrm{diag}(0,0,1,1)=\Lambda$$
Is this a typo in the authors' solution or am I not understanding this correctly?