Module over a Noncommutative Algebra and Taylor series

Let us consider the vector space $C^{\mathrm{\infty }}(\mathbb{R})$$C^{\infty}(\mathbb R)$, and some linear endomorphism on it:

$$\begin{array}{}\text{(1)}& \delta :f(x)⟼f^{\prime }(x),e:f(x)⟼f(a),t:f(x)⟼{\int }_a^{x}f(t)dt\end{array}$$

Here we view $f(a)$$f(a)$ as the constant function, hence $e$$e$ is an endomorphism.

Then consider the smallest $\mathbb{R}$$\mathbb R$ algebra contain $\delta ,e,t$$\delta,e,t$ in ${\mathrm{End}}_{\mathbb{R}-\mathrm{Mod}}(C^{\mathrm{\infty }}(\mathbb{R}))$$\mathrm{End}_{\mathbb R-\mathrm{Mod}}(C^{\infty}(\mathbb R))$, denote it as $(R,+,\ast )$$(R,+,*)$​​.

Notice that $R$$R$ is a Noncommutative ring since $\delta e=0,e\delta \ne 0$$\delta e=0,e\delta\ne 0$.

Now let us view $C^{\mathrm{\infty }}(\mathbb{R})$$C^{\infty}(\mathbb R)$ as a left module over $R$$R$.

By the fundamental theorem of Calculus, we have the identity

$$\begin{array}{}\text{(2)}& 1=e+t\delta \end{array}$$

i.e.

$$\begin{array}{}\text{(3)}& f(x)=f(a)+{\int }_a^x{f}^{\prime }(t)dt\end{array}$$

 

Since

$$\begin{array}{}\text{(4)}& f^{\prime }(t)=1\ast f^{\prime }(t)\end{array}$$

We have

$$\begin{array}{}\text{(5)}& f(x)=f(a)+{\int }_a^x(f^{\prime }(a)+{\int }_a^x{f}^{″}(t)dt)dt=f(a)+f^{\prime }(a)(x-a)+{\int }_a^x{\int }_a^x{f}^{″}(t)dt\end{array}$$

After using the identity $n+1$$n+1$ times we get

$$\begin{array}{}\text{(6)}& f(x)=\sum _{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a{)}^k+{\int }_a^x...{\int }_a^x{f}^{(n+1)}(t)dt\end{array}$$

By The First Mean Value Theorem for Integrals, we have

$$\begin{array}{}\text{(7)}& {\int }_a^x...{\int }_a^x{f}^{(n+1)}(t)dt={\int }_a^x...{\int }_a^x{f}^{(n+1)}(\zeta )dt=\frac{f^{(n+1)}(\zeta )}{(n+1)!}(x-a{)}^{n+1}\end{array}$$

Hence we get

$$\begin{array}{}\text{(8)}& f(x)=\sum _{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a{)}^k+\frac{f^{(n+1)}(\zeta )}{(n+1)!}(x-a{)}^{n+1}\end{array}$$