3
$\begingroup$

I am reading a paper entitled Bernoulli Numbers Via Determinants by Hongwei Chen and I'm confused about a particular step. The author sets up a system of equations via the following: first let $B_n$ represent the $n$-th Bernoulli number. Then

$$x=(e^x-1)\sum_{n=0}^\infty B_n\frac{x^n}{n!}$$

Letting $b_n=B_n/n!$ and expanding we get

$$x=\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\left(b_0+b_1x+b_2x^2+...\right)$$

Immediately we can see that $b_0=1$. Using Cauchy Products, we then obtain an infinite sequence of equations which are coefficients of the powers of $x$. But we don't need infinite, we can look at the system for coefficients up to $x^n$. Therefore

\begin{cases} b_1=-\frac{1}{2!} \\[2ex] \frac{b_1}{2!}+b_2=-\frac{1}{3!} \\[2ex] \frac{b_1}{3!}+\frac{b_2}{2!}+b_3=-\frac{1}{4!} \\[2ex] \vdots \\[2ex] \frac{b_1}{n!}+\frac{b_2}{(n-1)!}+...+b_n=-\frac{1}{(n+1)!} \\[2ex] \end{cases}

Then the author goes on to state that applying Cramers rule produces $$b_n= \begin{vmatrix} 1 & 0 & 0 & \cdots & -\frac{1}{2!} \\ \frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\ \end{vmatrix} $$

Then

$$B_n=n! \begin{vmatrix} 1 & 0 & 0 & \cdots & -\frac{1}{2!} \\ \frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\ \end{vmatrix} $$

And finally,

$$B_n=(-1)^n n! \begin{vmatrix} \frac{1}{2!} & 1 & 0 & \cdots & 0 \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & 0 \\ \frac{1}{4!} & \frac{1}{3!} & \frac{1}{2!} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{(n+1)!} &\frac{1}{n!} & \frac{1}{(n-1)!} & \cdots & \frac{1}{2!} \\ \end{vmatrix} $$

Can someone please explain the last step? I just don't see how that last step comes from the previous. I figure it has something to do with properties of determinants, and i do know that $|cA|=c^n|A|$, but I don't see where that is coming from here.

$\endgroup$
  • $\begingroup$ The last step comes from interchanging columns and $c=-1$. $\endgroup$ – Dietrich Burde Feb 15 '16 at 22:33
  • $\begingroup$ And so, if you do that, you only do that $n-1$ times...then the sign of the left most column is still negative...can you also take out a $-1$ then since the column is a scalar multiple? It's been a while since I've done determinants... $\endgroup$ – Iceman Feb 15 '16 at 22:34
  • $\begingroup$ @DietrichBurde, is my last statement correct? $\endgroup$ – Iceman Feb 15 '16 at 22:57
1
$\begingroup$

Here is the answer:

If $c$ is a constant and $A$ is an $n \times n$ matrix, then $\det c A = c^n \det A$. Note that here $c$ multiplies all columns not just the first column. Now, if $c$ multiplies only one column then $\det [A_1, A_2, \cdots, c A_i, \cdots A_n]= c \det A$.

Now for the problem. The constant $c=-1$ only multiplies the first column after being rolled from the last place ($n-1$ swaps).

Then we have the total sign is $(-1)^{n-1} (-1) = (-1)^n$.

$\endgroup$
  • 1
    $\begingroup$ In Mathematica, I typed in B[n_] := (-1)^n n! Det[ Table[If[r + 1 < c, 0, 1/(r - c + 2)!], {r, 1, n}, {c, 1, n}]] followed by BernoulliB[#] - B[#] & /@ Range[10]. The result was all zeros. $\endgroup$ – heropup Mar 11 '16 at 23:20
  • $\begingroup$ I see. So where is the error in Wolfram Alpha? $\endgroup$ – Herman Jaramillo Mar 11 '16 at 23:41
  • $\begingroup$ What Wolfram Alpha (and Mathematica) will do when you specify a matrix multiplied by a constant, is to multiply every element in the matrix by that constant. What you need to type is (wolframalpha.com/input/?i=4!+Det[{{1%2F2,1,0,0},{1%2F3!,1%2F2,1,0},{1%2F4!,1%2F3!,1%2F2,1},{1%2F5!,1%2F4!,1%2F3!,1%2F2}}]) instead. $\endgroup$ – heropup Mar 12 '16 at 0:03
  • $\begingroup$ @heropup I believe you are correct. I did it in maxima and it works fine. Thanks. $\endgroup$ – Herman Jaramillo Mar 12 '16 at 0:05
  • 1
    $\begingroup$ Yes, you could do that for rows or columns. The determinant of $A$ is equal to the determinant of $A^T$ (the transposed) so it does not matter. But there are two contributions to the sign. Swapping two columns changes the sign, and multiplying a column changes the sign by $(-1)^n$. Added together is $n+n-1 = 2n-1$ and for an odd number $ (-1)^{2n-1}=-1$ $\endgroup$ – Herman Jaramillo Mar 12 '16 at 0:22

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.