# Inverse of an Infinite Matrix (with factorials)

How to calculate this monstrous expression? $$\begin{pmatrix} \frac{1}{1!} & \frac{1}{2!} & \frac{1}{3!} & \frac{1}{4!} & \frac{1}{5!}& \cdots\\ 0 & \frac{1}{1!} & \frac{1}{2!} & \frac{1}{3!} & \frac{1}{4!}& \cdots \\ -2 & 0 & \frac{1}{1!} & \frac{1}{2!} & \frac{1}{3!} & \cdots \\ 0 & -3 & 0 & \frac{1}{1!} & \frac{1}{2!} & \cdots \\ 0 & 0 & -4 & 0 & \frac{1}{1!} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}^{-1} \begin{pmatrix}0\\1\\0\\0\\0\\0\\\vdots\end{pmatrix}$$ I don't think trying to find the inverse of this huge matrix (which I am not able to) will be helpful, as we only need the $2^{nd}$ column of the inverse matrix. Any help is appreciated! Thank You!

• Try doing the first steps of Gauss-Jordan elimination. I believe this will leave the 2nd column of the inverse unchanged after a finite number of steps. – Marcus Aurelius Jan 24 '18 at 13:34
• I tried doing it for 2 hours, but it did not stop. Probably, one would have to do it infinitely many times – Shashank Jan 24 '18 at 16:58

Using inversion by LDU-decomposition and including Euler-summation for the occuring divergent dot-products I get for the first couple of entries derivatives of the gamma-function $\Gamma(x)$ at argument $1$:

$$\begin{array}{r|rl} i & \text{num value} & \text{interpretation} \\ \hline 1 & -0.577215664902 & = \Gamma^{(1)} (1) \\ 2 & 1.97809833665 & = \Gamma^{(2)} (1) \\ 3 & -5.44487445649 & = \Gamma^{(3)} (1) \\ 4 & 23.5614740841 & = \Gamma^{(4)} (1) \end{array}$$

So I think, this continues for the other entries of the result-vector and the interpretation, suggested by the approximations, hold in general.

Using Pari/GP we get the found values in the exponential generating-function for the $\Gamma(1+x)$:

serlaplace(gamma(1+x)-1)
%168 = -0.577215664902*x + 1.97811199066*x^2
- 5.44487445649*x^3 + 23.5614740840*x^4
- 117.839408268*x^5 + 715.067362527*x^6
- 5019.84887263*x^7 + 40243.6215733*x^8
- 362526.289115*x^9 + 3627042.41276*x^10
- 39907084.1514*x^11 + 478943291.765*x^12
- 6226641351.55*x^13 + 87175633810.7*x^14
- 1.30765442950 E12*x^15 + O(x^16)

Appendix Tables

Here are the top-left segments of the LDU-components such that $M=L \cdot D \cdot U$:

1    .      .      .       .        .  |
.    1      .      .       .        .  |
-2    1      1      .       .        .  |
.   -3    9/5      1       .        .  |      L
.    .  -24/5    8/3       1        .  |
.    .      .  -20/3  185/52        1  |
-    -      -      -       -        -  +
1    .      .      .       .        .  |
.    1      .      .       .        .  |
.    .    5/6      .       .        .  |      D
.    .      .    3/4       .        .  |
.    .      .      .   52/75        .  |
.    .      .      .       .  203/312  |
-    -      -      -       -        -  +
1  1/2    1/6   1/24   1/120    1/720  |
.    1    1/2    1/6    1/24    1/120  |
.    .      1    1/2  17/100   13/300  |       U
.    .      .      1   37/75   38/225  |
.    .      .      .       1  151/312  |
.    .      .      .       .        1  |
-    -      -      -       -        -  +

Here are their inverses, such that $$M^{-1} = \lim_{dim \to \infty} U^{-1} \underset {\mathfrak E} * ( D^{-1} \cdot L^{-1})$$ where $\underset {\mathfrak E} *$ means doing the divergent dotproducts using Euler-summation

1     -1/2     1/12       .   -1/600  -1/37440  |
.        1     -1/2    1/12    1/450  -17/9360  |
.        .        1    -1/2   23/300    5/1248  |
.        .        .       1   -37/75  109/1560  |    U^-1
.        .        .       .        1  -151/312  |
.        .        .       .        .         1  |
-        -        -       -        -         -  +
1        .        .       .        .         .  |
.        1        .       .        .         .  |
.        .      6/5       .        .         .  |
.        .        .     4/3        .         .  |    D^-1
.        .        .       .    75/52         .  |
.        .        .       .        .   312/203  |
-        -        -       -        -         -  +
1        .        .       .        .         .  |
.        1        .       .        .         .  |
2       -1        1       .        .         .  |    L^-1
-18/5     24/5     -9/5       1        .         .  |
96/5    -88/5     48/5    -8/3        1         .  |
-1200/13  1230/13  -600/13  210/13  -185/52         1  |
-        -        -       -        -         -  +

Because we have the convergent dotproduct $L^{-1} \cdot I_1$ where $I_1 =[0,1,0,0,...]$ we need only the second column of $L^{-1}$ and the part $\text{rhs}=D^{-1} \cdot L^{-1} \cdot I_1$ gives, in decimal notation

.  |
1.00000000000  |
-1.20000000000  |
6.40000000000  |
-25.3846153846  |
145.418719212  |
-930.992018244  |   rhs=D^-1 * L^-1 * I[,1]
6963.47826087  |  (dotproducts of finite lengthes are exact)
-58772.5918570  |
554512.555292  |
-5779721.10527  |
65970421.8290  |
....           |

The dot-products of left-multiplication with $U^{-1}$ include Eulersummation to assign the divergent sums of alternating series finite values. We get the following approximations:

-0.577218921840  |
1.97813788259  |
-5.44460173003  | fairly good aproximations to coefficients of the
23.5552716067  | laplace-transformation of the Gamma(1+x)-power series
-117.723592387  | see above
712.587241686  |
(-4968.35894817) |  approximation at higher indexes worsen because
( ...   )        |  of finite size of matrices