Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/95

when ${\text{N}},$ which is a whole number, is increased indefinitely. That this definition is equivalent to the preceding, will appear if the expression is expanded by the binomial theorem, which is evidently applicable in a case of this kind.

These functions of $\Phi$ are homologous with $\Phi .$

172. We may define the logarithm as the function which is the inverse of the exponential, so that the equations

${\begin{aligned}e^{\Psi }&=\Phi ,\\\Psi &=\log \Phi ,\end{aligned}}$

are equivalent, leaving it undetermined for the present whether every dyadic has a logarithm, and whether a dyadic can have more than one.

173. It follows at once from the second definition of the exponential function that, if $\Phi$ and $\Psi$ are homologous,

$e^{\Phi }.e^{\Psi }=e^{\Phi +\Psi },$

and that, if ${\text{T}}$ is a positive or negative whole number,

$\{e^{\Phi }\}^{\text{T}}=e^{{\text{T}}\Phi }.$

174. If $\Xi$ and $\Phi$ are homologous dyadics, and such that

$\Xi ^{2}.\Phi =-\Phi ,$

the definitions of No. 171 give immediately

${\begin{aligned}e^{\Xi .\Phi }&=\cos \Phi +\Xi \,\sin \Phi ,\\e^{-\Xi .\Phi }&=\cos \Phi -\Xi \,\sin \Phi ,\\\end{aligned}}$

whence

${\begin{aligned}\cos \Phi &={\tfrac {1}{2}}\{e^{\Xi \Phi }+e^{-\Xi \Phi }\},\\\sin \Phi &=-{\tfrac {1}{2}}\Xi \{e^{\Xi \Phi }-e^{-\Xi \Phi }\}.\end{aligned}}$

175. If $\Phi .\Psi =\Psi .\Phi =0,$

$\{\Phi +\Psi \}^{2}=\Phi ^{2}+\Psi ^{2},$$\{\Phi +\Psi \}^{3}=\Phi ^{3}+\Psi ^{3},{\text{ etc.}}$

Therefore

${\begin{aligned}e^{\Phi +\Psi }&=e^{\Phi }+e^{\Psi }-{\text{I}},\\\cos\{\Phi +\Psi \}&=\cos \Phi +\cos \Psi -{\text{I}},\\\sin\{\Phi +\Psi \}&=\sin \Phi +\sin \Psi .\end{aligned}}$

176.

$\left\vert e^{\Phi }\right\vert =e^{\Phi _{\text{S}}}.$

For the first member of this equation is the limit of

$\left\vert \{{\text{I}}+{\text{N}}^{-1}\Phi \}^{\text{N}}\right\vert ,$that is, of$\left\vert {\text{I}}+{\text{N}}^{-1}\Phi \right\vert ^{\text{N}}.$

If we set $\Phi =\alpha i+\beta j+\gamma k,$ the limit becomes that of

$(1+{\text{N}}^{-1}\alpha .i+{\text{N}}^{-1}\beta .j+{\text{N}}^{-1}\gamma .k)^{\text{N}},$or$(1+{\text{N}}^{-1}\Phi _{\text{S}})^{\text{N}},$

the limit of which is the second member of the equation to be proved.

177. By the definition of exponentials, the expression

$e^{q\{kj-jk\}}$

represents the limit of

$\{{\text{I}}+q{\text{N}}^{-1}\{kj-jk\}\}^{\text{N}}.$