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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 $O\rightarrow \epsilon^3$ and expand in $\epsilon$ again.

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

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Well $A = \sum_{i=0}^\infty a_i\varepsilon^i$ is invertible as formal power series in $\varepsilon$ so long as $a_0$ is invertible. In this case if the inverse is $B = \sum_{i=0}^\infty b_i\varepsilon^i$ then by writing $\sum_{i=0}^\infty c_i\varepsilon^i = AB$, 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_1\varepsilon + a_2\varepsilon^2 + \cdots)(b_0+b_1\varepsilon + b_2\varepsilon^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:

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

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_0\bigg(\sum_{n = i_1+\cdots + i_r\\1\leq 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_{1\leq i_1 \leq n} a_{i_1}\bigg(\sum_{n - i_1 = i_2+\cdots + i_r\\1\leq 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\\ 1\leq 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\\ 1\leq 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.

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  • $\begingroup$ Could you please add a link to a proof for the coefficients' inverse formula? $\endgroup$ – usumdelphini Jan 11 at 14:48
  • $\begingroup$ @usumdelphini Sure, added. $\endgroup$ – Ben Jan 11 at 15:49
  • $\begingroup$ Thanks, can you please have a look at this as well? math.stackexchange.com/questions/3075846/… $\endgroup$ – usumdelphini Jan 16 at 15:15
  • $\begingroup$ @usumdelphini Since the collection of doubly infinite series no longer forms a ring (you can't multiply), this is not a formal algebraic problem in the same way. There may not be an inverse anymore, you actually have to think about convergence conditions on sums of coefficients, and this makes the problem much harder. I don't have anything to add there. $\endgroup$ – Ben Jan 16 at 15:36

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