Linear map of \(\varvec{D}\) -algebra \(\varvec{A}\) is linear map of \(\varvec{D}\) -module \(\varvec{A}\) . We use notation \( f \circ a=f(a) \) for image of the linear map \(\varvec{f}\) . We can identify linear map \(\varvec{f}\) of \(\varvec{D}\) -algebra \(\varvec{A}\) and tensor \(f \in A^{2 \otimes }\) according to the equality \( (a \otimes b) \circ c=a c b \) I considered solving equation \( a \circ x=b \) Let \(\varvec{A}\) be Banach \(\varvec{D}\) -algebra. The map \( f: A \rightarrow A \) is called differentiable on the set \(\varvec{U} \subset \varvec{A}\) , if at every point \(\varvec{x} \in \varvec{U}\) the increment of map \(\varvec{f}\) can be represented as where is linear map and \( o: A \rightarrow A \) is such continuous map that \( \lim _{h \rightarrow 0} \frac{\Vert o(h)\Vert }{\Vert h\Vert }=0 \) Newton’s method to solve the equation \( f(x)=a \) where \(\varvec{f}\) is differentiable map is iterative method for finding the root. Let \(\varvec{x}_{\textbf{0}}\) be initial approximation for the step of iteration. Then the solution of linear equation \( \frac{df(x)}{dx} \circ x=-f\left( x_{0}\right) +\frac{df(x)}{dx} \circ x_{0} \) is initial approximation for the next step of iteration.
