Dtyjlui

Let's assume I want to invert a matrix function $M=M(x)$, which is expressed as a power series of the small parameter $epsilon$

$$M = M_0(x) + M_1(x) epsilon + M_2(x) epsilon^2 + mathcal{O}(epsilon^3)$$

I would like to invert this and obtain $M^{-1}(x)$ as a series expansion itself.

Option 1 - The temptation is to invert a truncated version of $M$, but this could cause distortions, due to the possible role of the terms $mathcal{O}(epsilon^3)$.

Option 2 - I could also truncate $M$ and add to it a placeholder matrix $O$, to symbolize all the terms $mathcal{O}(epsilon^3)$. When I obtain the inverse, it will be in terms of $x$ and $O$. I could then substitute $Orightarrow epsilon^3$ and expand in $epsilon$ again.

None of these strike me as good methods. What would be the best way to proceed?

Well $A = sum_{i=0}^infty a_ivarepsilon^i$ is invertible as formal power series in $varepsilon$ so long as $a_0$ is invertible. In this case if the inverse is $B = sum_{i=0}^infty b_ivarepsilon^i$ then by writing $sum_{i=0}^infty c_ivarepsilon^i = AB$, solving for the coefficients $c_i$ and setting $c_i = delta_{i,0}$ you will get coefficients. The first few terms for the inverse are:

$$b_0 = a_0^{-1},$$
$$b_1 = -a_0^{-1}a_1a_0^{-1}$$
$$b_2 = a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} -a_0^{-1}a_2a_0^{-1}$$
$$b_3 = - a_0^{-1}a_1a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} + a_0^{-1}a_1a_0^{-1}a_2a_0^{-1} + a_0^{-1}a_2a_0^{-1}a_1a_0^{-1} - a_0^{-1}a_3a_0^{-1}$$

and I imagine you see the pattern by now (the formula is a lot easier to read if $a_0 = 1$).

However, this won't work for your function as stated because it appears the constant term is zero. Did you mean to write:

$$M(x+varepsilon) = M_0(x) + M_1(x)varepsilon + M_2(x)varepsilon^2 + O(varepsilon^3)$$

or do you really assume $M_0(x) = 0$?

If $M_0 = 0$ then $M(x+varepsilon)$ is not invertible at $varepsilon = 0$, but you can find an inverse in the punctured neighborhood around $x$, i.e. with Laurent series. If you assume $M_1$ is invertible then you can do the same thing as above to get an inverse for $varepsilon^{-1}M(x+varepsilon)$ given by:
$$M_1^{-1}(x)varepsilon^{-1} - M_1(x)^{-1}M_2(x)M_1(x)^{-1} + (cdots)varepsilon + (cdots)varepsilon^2 + O(varepsilon^3)$$
where the linear and quadratic terms are as in $b_2,b_3$ above.

ADDED:

There is no proof anywhere for the coefficients, but here is how you get them.

begin{align}
1 &= (a_0+a_1varepsilon + a_2varepsilon^2 + cdots)(b_0+b_1varepsilon + b_2varepsilon^2 + cdots)\
&= a_0b_0 + (a_1b_0 + a_0b_1)varepsilon + (a_2b_0 + a_1b_1 + a_0b_2)varepsilon^2 + (a_3b_0 + a_2b_1 + a_1b_2 + a_0b_3)varepsilon^3 + cdots
end{align}

Now:

• Set $a_0b_0 = 1$ to get $b_0 = a_0^{-1}$.




• All the higher terms have to be zero, so set $a_1b_0 + a_0b_1 = 0$ and substitute the value of $b_0$ to get $a_1a_0^{-1} + a_0b_1 = 0$. Now solve for $b_1$.




• Plug $b_0,b_1$ into the next equation $a_2b_0 + a_1b_1 + a_0b_2 = 0$ and solve.




• Keep going with this as long as you want.

Or you can prove that the formula looks a certain way:

$$b_n = sum_{n = i_1+cdots + i_r\1leq i_1,ldots, i_r} (-1)^ra_0^{-1}a_{i_1}a_0^{-1}a_{i_2}a_0^{-1}cdots a_0^{-1}a_{i_r}a_0^{-1}$$

You have to see that this solution satisfies $sum_{i=0}^n a_ib_{n-i} = 0$ for $n geq 1$. It seems more or less clear, but I guess it would go something like this:

begin{align}
a_0b_n &= a_0bigg(sum_{n = i_1+cdots + i_r\1leq i_1,ldots, i_r} (-1)^ra_0^{-1}a_{i_1}a_0^{-1}a_{i_2}a_0^{-1}cdots a_0^{-1}a_{i_r}a_0^{-1}bigg)\
&= sum_{1leq i_1 leq n} a_{i_1}bigg(sum_{n - i_1 = i_2+cdots + i_r\1leq i_2,ldots, i_r} (-1)^ra_0^{-1}a_{i_2}a_0^{-1}a_{i_3}a_0^{-1}cdots a_0^{-1}a_{i_r}a_0^{-1}bigg)\
&= sum_{j=1}^n a_{j}bigg(sum_{n - j = j_1+cdots + j_k\ 1leq j_1,ldots, j_k} (-1)^{k-1}a_0^{-1}a_{j_1}a_0^{-1}a_{j_2}a_0^{-1}cdots a_0^{-1}a_{j_k}a_0^{-1}bigg)\
&= sum_{j=1}^n(-1)a_{j}bigg(sum_{n - j = j_1+cdots + j_k\ 1leq j_1,ldots, j_k} (-1)^{k}a_0^{-1}a_{j_1}a_0^{-1}a_{j_2}a_0^{-1}cdots a_0^{-1}a_{j_k}a_0^{-1}bigg)\
&= (-1)sum_{j=1}^n a_jb_{n-j}
end{align}

Done. Moving to the second line I factored out $a_0^{-1}a_{i_1}$, to the next line reindexed all the sums, to the next line factored out $(-1)$, to the next line substituted the $b$ coefficients.