# proof of Taylor’s formula for matrix functions

###### Theorem.

Let $p$ be a polynomial and suppose $\mathbf{A}$ and $\mathbf{B}$ are squared matrices of the same size, then ${\displaystyle p(\mathbf{A}+\mathbf{B})=\sum_{k=0}^{n}\frac{1}{k!}p^{(k)}(% \mathbf{A})\mathbf{B}^{k}}$ where $n=\deg(p)$.

###### Proof.

Since $p$ is a polynomial, we can apply the Taylor expansion:

 $p(x)=\sum_{k=0}^{n}\frac{1}{k!}p^{\left(k\right)}\left(x_{0}\right)\left(x-x_{% 0}\right)^{k}$

where $n=\deg(p)$. Now let $x=\mathbf{A}+\mathbf{B}$ and $x_{0}=\mathbf{A}$.

The Taylor expansion can be checked as follows: let $p\left(x\right)=\sum_{k=0}^{n}a_{k}x^{k}$ for coefficients $a_{k}$ (note that this coefficients can be taken from the space of square matrices defined over a field). We define the formal derivative of this polynomial as $p^{\left(1\right)}\left(x\right)=\frac{dp}{dx}=\sum_{k=1}^{n}a_{k}kx^{k-1}$ and we define $p^{\left(k\right)}=\frac{dp^{\left(k-1\right)}}{dx}$.

Then $p^{\left(k\right)}\left(x\right)=\sum_{i=k}^{n}a_{i}\frac{i!}{\left(i-k\right)% !}x^{i-k}$ and we have $\frac{1}{k!}p^{\left(k\right)}\left(x_{0}\right)=\sum_{i=k}^{n}a_{i}\frac{i!}{% \left(i-k\right)!k!}\left(x_{0}\right)^{i-k}$. Now consider

 $\displaystyle\sum_{k=0}^{n}\frac{1}{k!}p^{\left(k\right)}\left(x_{0}\right)% \left(x-x_{0}\right)^{k}=\sum_{k=0}^{n}\left(\sum_{i=k}^{n}a_{i}\frac{i!}{% \left(i-k\right)!k!}\left(x_{0}\right)^{i-k}\left(x-x_{0}\right)^{k}\right)$ $\displaystyle=$ $\displaystyle\sum_{i=0}^{n}a_{i}\left(x_{0}\right)^{i}+\sum_{i=1}^{n}a_{i}i% \left(x_{0}\right)^{i-1}\left(x-x_{0}\right)+\dots+\sum_{i=j}^{n}a_{i}\frac{i!% }{\left(i-j\right)!j!}\left(x_{0}\right)^{i-j}\left(x-x_{0}\right)^{j}+\dots+a% _{n}\left(x-x_{0}\right)^{n}$ $\displaystyle=$ $\displaystyle a_{0}+a_{1}\left(x\right)+\dots+a_{i}\left(\sum_{j=0}^{i}\frac{i% !}{\left(i-j\right)!j!}\left(x_{0}\right)^{i-j}\left(x-x_{0}\right)^{j}\right)% +\dots+a_{n}\left(\sum_{j=0}^{n}\frac{n!}{\left(n-j\right)!j!}\left(x_{0}% \right)^{n-j}\left(x-x_{0}\right)^{j}\right)$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{n}a_{k}x^{i}=p\left(x\right)$

since $\sum_{j=0}^{i}\frac{i!}{\left(i-j\right)!j!}\left(x_{0}\right)^{i-j}\left(x-x_% {0}\right)^{j}=\left(x\right)^{i}$. ∎

Title proof of Taylor’s formula for matrix functions ProofOfTaylorsFormulaForMatrixFunctions 2013-03-22 17:57:04 2013-03-22 17:57:04 joen235 (18354) joen235 (18354) 4 joen235 (18354) Proof msc 47A56