2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Marcus Aurelius Jan 24 '18 at 13:34
  • $\begingroup$ I tried doing it for 2 hours, but it did not stop. Probably, one would have to do it infinitely many times $\endgroup$ – Shashank Jan 24 '18 at 16:58
0
$\begingroup$

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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.