12
$\begingroup$

I have to find this determinant, call it $D$ \begin{vmatrix} \frac12 & \frac1{3}& \frac1{4} & \dots & \frac1{n+1} \\ \frac1{3} & \frac14 & \frac15 & \dots & \frac1{n+2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac1{n+1} & \frac1{n+2} & \frac1{n+3} & \dots & \frac1{2n} \end{vmatrix}

As there are no zeros in there, despite being a symmetric matrix, finding this determinant is tough for me. Is there any tricks etc. I do not know any softwares to find this determinant.


I tried to make a pattern by calculating for $n=1,2,3,\dots$

$n=1 \hspace{5cm}D_1=\frac12\\ n=2 \hspace{5cm} D_2=\frac12\frac14-\frac13\frac13\\ \\n=3 \hspace{5cm} D_3=\frac12\frac14\frac16-\frac13\frac13\frac16-\frac12\frac15\frac15-\frac14\frac14\frac14+\frac13\frac14\frac15+\frac13\frac14\frac15$

What I spot from here is, to get $D_2$, (even case)

We get $\frac12\frac14$ by multiplying diagonal entries and then subtracting $\frac1x\frac1x$ where $x=\frac{2+4}2$

Next, to get $D_3$ (odd case), we get our first term i.e. $\frac12\frac14\frac16$ by multiplying diagonal entries and to obtain the rest we follow this pattern that subtract $\frac1x\frac1y\frac1z$, where first we fix $x=2$ and make $y=z=\frac{4+6}2=5$, similarly next we subtract by fixing $x=6$ and $y=z=\frac{2+4}2=3$ and then by fixing $x=4$ and $y=z=\frac{4+4}2=4$, and to get terms that get added we add terms of the form $\frac1x\frac1y\frac1z$ by putting $x=\frac{2+4}2 , y=\frac{4+4}2=4,z= \frac{4+6}2=5$, but we do it $2$ times.

Now out of curiosity I had to calculate $n=4,5$. Here are them-

For $n=4 \text{(even case)} \hspace{3cm} D_4=\frac1{2.4.6.8}-\frac1{2.5.5.8}-\frac1{2.4.7.7}-\frac1{2.6.6.6}\frac1{3.3.6.8}-\frac1{3.4.6.7}-\frac1{5.5.3.7}-\frac1{3.4.6.7}-\frac1{4.4.4.8}-\frac1{4.5.5.6}-\frac1{3.5.5.7}-\frac1{4.5.5.6}-\frac1{4.5.5.6}+\frac1{2.5.6.7}+\frac1{2.5.6.7}+\frac1{4.5.5.6}+\frac1{3.3.7.7}+\frac1{3.4.5.8}+\frac1{3.6.6.5}+\frac1{3.4.5.8}+\frac1{4.4.5.7}+\frac1{4.4.6.6}+\frac1{3.6.6.5}+\frac1{5.5.5.5}+\frac1{4.4.5.7}$.

This helps somewhat in recognizing a pattern in even case, but to be sure one has to find $n=6$ case too.

I guess $n=5$ case will be enough to recognize a pattern, if the one above mentioned in $n=3$ works, as by that, $D_5$ should come out to be

$D_5=\frac1{2.4.6.8.10}- {^5C_2} \text{terms of the form} \frac1{x_1x_2x_3x_4x_5}, \text{where any three terms say} x_1,x_2,x_3$ are fixed out of $2,4,6,8,10$ and $x_4,x_5$ takes values of mean of the other two remaining and similarly for positive terms, but they seem very less number of terms as there were already $12$ negative terms in expansion of $D_4$, so here some more negative terms will appear and their pattern can be known by only finding them, after a few steps, may be upto $n=11$ or $12$, we can see a general pattern appearing.

I am sure after calculating all this that there is a pattern, but may be too complex to find by hand, as calculations gets huge, and it is also probably a hammer to kill an ant, as I am not aware of any other trick, I worked all this out, may be someone can find a quicker solution?

$\endgroup$
  • $\begingroup$ This is an example of the Hankel matrix: en.wikipedia.org/wiki/Hankel_matrix $\endgroup$ – DVD Jul 4 '15 at 21:57
  • $\begingroup$ Replace $c_{i,j} = \frac{1}{i+j}$ in the $(i,j)$-th position of the matrix with the more general scenario $c'_{i,j} = \frac{1}{x_i+x_j}$ and see if you can show $\displaystyle \det (c'_{i,j}) = \prod\limits_{1 \le i < j \le n} (x_i - x_j)^2 \prod\limits_{1 \le i \le j \le n} \frac{1}{x_i + x_j}$ .. $\endgroup$ – r9m Jul 4 '15 at 23:06
  • $\begingroup$ Related: math.stackexchange.com/questions/1316568/… $\endgroup$ – Martin Sleziak Feb 29 '16 at 13:47
9
$\begingroup$

I guess it's easier to go for a reduction formula. I proceed along the generalization mentioned in comment:

Call the determinant $M = M_n(x_1,x_2,\cdots, x_n)$

$$\displaystyle \begin{align}M &= \left|\begin{matrix}\dfrac{1}{x_1+x_1} & \cdots & \dfrac{1}{x_1+x_n}\\ \dfrac{1}{x_2+x_1} & \cdots & \dfrac{1}{x_2+x_n}\\ \cdot & \cdot &\cdot \\ \dfrac{1}{x_n+x_1} & \cdots & \dfrac{1}{x_n+x_n}\end{matrix}\right| \\& \text{subtract R1 from each row} \\&= \left|\begin{matrix}\dfrac{1}{x_1+x_1} & \cdots & \dfrac{1}{x_1+x_n}\\ \dfrac{x_1 - x_2}{(x_2+x_1)(x_1+x_1)} & \cdots & \dfrac{x_1 - x_2}{(x_2+x_n)(x_1+x_n)}\\ \cdot & \cdot &\cdot \\ \dfrac{x_1 - x_n}{(x_n+x_1)(x_1+x_1)} & \cdots & \dfrac{x_1 - x_n}{(x_n+x_n)(x_1+x_n)}\end{matrix}\right| \\&= \prod\limits_{j=2}^{n} (x_1 - x_j)\times \prod\limits_{j=1}^{n}\frac{1}{(x_1+x_j)} \times \left|\begin{matrix}1 & \cdots & 1\\ \dfrac{1}{(x_2+x_1)} & \cdots & \dfrac{1}{(x_2+x_n)}\\ \cdot & \cdot &\cdot \\ \dfrac{1}{(x_n+x_1)} & \cdots & \dfrac{1}{(x_n+x_n)}\end{matrix}\right|\end{align}$$

Again subtracting $\textrm{C1}$ from each column,

$$\begin{align}\left|\begin{matrix}1 & \cdots & 1\\ \dfrac{1}{(x_2+x_1)} & \cdots & \dfrac{1}{(x_2+x_n)}\\ \cdot & \cdot &\cdot \\ \dfrac{1}{(x_n+x_1)} & \cdots & \dfrac{1}{(x_n+x_n)}\end{matrix}\right| &= \left|\begin{matrix}1 & \cdots & 0\\ \dfrac{1}{(x_2+x_1)} & \cdots & \dfrac{x_1 - x_n}{(x_2+x_n)(x_2+x_1)}\\ \cdot & \cdot &\cdot \\ \dfrac{1}{(x_n+x_1)} & \cdots & \dfrac{x_1 - x_n}{(x_n+x_n)(x_n+x_1)}\end{matrix}\right|\\&= \prod\limits_{j=2}^{n} (x_1-x_j) \times \prod\limits_{j=2}^n \frac{1}{x_j+x_1} \times \left|\begin{matrix}1 & \cdots & 0\\ 1 & \cdots & \dfrac{1}{(x_2+x_n)}\\ \cdot & \cdot &\cdot \\ 1 & \cdots & \dfrac{1}{(x_n+x_n)}\end{matrix}\right|\\&= \prod\limits_{j=2}^{n} (x_1-x_j) \times \prod\limits_{j=2}^n \frac{1}{x_j+x_1} \times M_{n-1}(x_2,x_3,\cdots ,x_n)\end{align}$$

Hence, $$M_n(x_1,x_2,\cdots, x_n) = \frac{\prod\limits_{1 \le j < i \le n} (x_i-x_j)^2}{\prod\limits_{1 \le i,j \le n} (x_i+x_j)}$$

$\endgroup$
  • 1
    $\begingroup$ This is the most straightforward way. $\endgroup$ – Gabriel Romon Jul 5 '15 at 7:24
5
$\begingroup$

We consider a general case. Let $$ |A|=\left|\begin{array}{}{\dfrac1{a_1+b_1}\cdots\dfrac1{a_1+b_n}\\ \vdots\hspace{20 mm} \vdots \\\dfrac1{a_n+b_1}\cdots\dfrac1{a_n+b_n}} \end{array}\right| $$ In your case, just set $a_i=i,b_j=j$.

By multiplying $i^{th}$ row with $\prod\limits_{j=1}^{n}(a_i+b_j)$ each, we have $$ |A|=\dfrac1{\prod\limits_{i,j=1}^{n}(a_i+b_j)}\left|\begin{array}{}{\prod\limits_{j=2}^{n}(a_1+b_j)\cdots\prod\limits_{j=1}^{n-1}(a_1+b_j)\\ \vdots\hspace{20 mm} \vdots \\\prod\limits_{j=2}^{n}(a_n+b_j)\cdots\prod\limits_{j=1}^{n-1}(a_n+b_j)} \end{array}\right| \hspace{20 mm}(1) $$ Since each $(i,j)$ element in $(1)$ can be expanded into polynomial, i.e $$ \prod\limits_{k=1,k\ne j}^{n}(a_i+b_k)=a_i^{n-1}+\sum\limits_{k=1,k\ne j}^{n}{b_k}a_i^{n-2}+\cdots+\prod\limits_{k=1,k\ne j}^{n}{b_k}=\pmatrix{1&a_i&\cdots &a_i^{n-1}}\pmatrix{\prod\limits_{k=1,k\ne j}^{n}{b_k}\\ \vdots \\ \sum\limits_{k=1,k\ne j}^{n}{b_k}\\1} $$ So \begin{align} \left|\begin{array}{}{\prod\limits_{j=2}^{n}(a_1+b_j)\cdots\prod\limits_{j=1}^{n-1}(a_1+b_j)\\ \vdots\hspace{20 mm} \vdots \\\prod\limits_{j=2}^{n}(a_n+b_j)\cdots\prod\limits_{j=1}^{n-1}(a_n+b_j)} \end{array}\right|&=\left|\begin{array}{}{1 \hspace{4 mm} a_1 \hspace{4 mm}\cdots\hspace{4 mm}a_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} a_n \hspace{4 mm}\cdots\hspace{4 mm}a_n^{n-1}} \end{array}\right|\left|\begin{array}{}{\prod\limits_{k=1,k\ne 1}^{n}{b_k} \hspace{4 mm} \prod\limits_{k=1,k\ne 2}^{n}{b_k} \hspace{4 mm}\cdots\hspace{4 mm}\prod\limits_{k=1,k\ne n}^{n}{b_k}\\ \vdots\hspace{40 mm} \vdots \\ \sum\limits_{k=1,k\ne 1}^{n}{b_k} \hspace{4 mm} \sum\limits_{k=1,k\ne 2}^{n}{b_k} \hspace{4 mm}\cdots\hspace{4 mm}\sum\limits_{k=1,k\ne n}^{n}{b_k}\\1\hspace{20 mm}1\hspace{7 mm}\cdots\hspace{7 mm} 1} \end{array}\right| \\ &=\left|\begin{array}{}{1 \hspace{4 mm} a_1 \hspace{4 mm}\cdots\hspace{4 mm}a_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} a_n \hspace{4 mm}\cdots\hspace{4 mm}a_n^{n-1}} \end{array}\right|f(b_1\cdots b_n) \\ &=R_1(b_1\cdots b_n)\tag{2} \end{align} where $f(b_1\cdots b_n)$ is a polynomial of $b_1\cdots b_n$.

Now by multiplying $j^{th}$ column with $\prod\limits_{i=1}^{n}(a_i+b_j)$ each, we have $$ |A|=\dfrac1{\prod\limits_{i,j=1}^{n}(a_i+b_j)}\left|\begin{array}{}{\prod\limits_{i=2}^{n}(a_i+b_1)\cdots\prod\limits_{i=1}^{n-1}(a_i+b_1)\\ \vdots\hspace{20 mm} \vdots \\\prod\limits_{i=2}^{n}(a_i+b_n)\cdots\prod\limits_{i=1}^{n-1}(a_i+b_n)} \end{array}\right| \hspace{20 mm}(3) $$

Expand each $(i,j)$ element in $(3)$ into polynomial of $b_j$, we have $$ \prod\limits_{k=1,k\ne i}^{n}(a_k+b_j)=b_j^{n-1}+\sum\limits_{k=1,k\ne i}^{n}{a_k}b_j^{n-2}+\cdots+\prod\limits_{k=1,k\ne i}^{n}{a_k}=\pmatrix{1&b_j&\cdots &b_j^{n-1}}\pmatrix{\prod\limits_{k=1,k\ne i}^{n}{a_k}\\ \vdots \\ \sum\limits_{k=1,k\ne i}^{n}{a_k}\\1} $$ So \begin{align} \left|\begin{array}{}{\prod\limits_{i=2}^{n}(a_i+b_1)\cdots\prod\limits_{i=1}^{n-1}(a_i+b_1)\\ \vdots\hspace{20 mm} \vdots \\\prod\limits_{i=2}^{n}(a_i+b_n)\cdots\prod\limits_{i=1}^{n-1}(a_i+b_n)} \end{array}\right| &=\left|\begin{array}{}{1 \hspace{4 mm} b_1 \hspace{4 mm}\cdots\hspace{4 mm}b_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} b_n \hspace{4 mm}\cdots\hspace{4 mm}b_n^{n-1}} \end{array}\right|\left|\begin{array}{}{\prod\limits_{k=1,k\ne 1}^{n}{a_k} \hspace{4 mm} \prod\limits_{k=1,k\ne 2}^{n}{a_k} \hspace{4 mm}\cdots\hspace{4 mm}\prod\limits_{k=1,k\ne n}^{n}{a_k}\\ \vdots\hspace{40 mm} \vdots \\ \sum\limits_{k=1,k\ne 1}^{n}{a_k} \hspace{4 mm} \sum\limits_{k=1,k\ne 2}^{n}{a_k} \hspace{4 mm}\cdots\hspace{4 mm}\sum\limits_{k=1,k\ne n}^{n}{a_k}\\1\hspace{20 mm}1\hspace{7 mm}\cdots\hspace{7 mm} 1} \end{array}\right| \\ &=\left|\begin{array}{}{1 \hspace{4 mm} b_1 \hspace{4 mm}\cdots\hspace{4 mm}b_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} b_n \hspace{4 mm}\cdots\hspace{4 mm}b_n^{n-1}} \end{array}\right|f(a_1\cdots a_n) \\ &=R_2(b_1\cdots b_n)\tag{4} \end{align}

Now consider $R_1-R_2$ as polynomial of $b_1\cdots b_n$. Since the number of roots of $R_1-R_2$ is more than its degree, $R_1-R_2$ vanishes and the coefficients of $b_1\cdots b_n$ in $R_1$ must be equal to those of $R_2$. By comparing $(2)$ and $(4)$, we have $$ f(b_1\cdots b_n)=\left|\begin{array}{}{1 \hspace{4 mm} b_1 \hspace{4 mm}\cdots\hspace{4 mm}b_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} b_n \hspace{4 mm}\cdots\hspace{4 mm}b_n^{n-1}} \end{array}\right| $$ So from $(1)$ and $(2)$, we have \begin{align} |A|&=\dfrac1{\prod\limits_{i,j=1}^{n}(a_i+b_j)}\left|\begin{array}{}{1 \hspace{4 mm} a_1 \hspace{4 mm}\cdots\hspace{4 mm}a_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} a_n \hspace{4 mm}\cdots\hspace{4 mm}a_n^{n-1}} \end{array}\right|\left|\begin{array}{}{1 \hspace{4 mm} b_1 \hspace{4 mm}\cdots\hspace{4 mm}b_1^{n-1}\\ \vdots\hspace{20 mm} \vdots \\ 1 \hspace{4 mm} b_n \hspace{4 mm}\cdots\hspace{4 mm}b_n^{n-1}} \end{array}\right| \\ &=\dfrac{\prod\limits_{1\le i<j\le n} (a_j-a_i)\prod\limits_{1\le i<j\le n} (b_j-b_i)}{\prod\limits_{i,j=1}^{n}(a_i+b_j)} \end{align} In your case $$ |A|=\dfrac{\prod\limits_{1\le i<j\le n} (j-i)^2}{\prod\limits_{i,j=1}^{n}(i+j)} $$

$\endgroup$
  • $\begingroup$ How does "comparing (2) and (4)" work in your proof? What if $f $ was not the Vandermonde determinant but rather some scalar multiple thereof? $\endgroup$ – darij grinberg Nov 1 at 6:12
  • $\begingroup$ If you have two polynomials $f$ and $g$ (each in $n$ variables) and you know that $f\left(a_1,a_2,\ldots,a_n\right) g\left(b_1,b_2,\ldots,b_n\right) = f\left(b_1,b_2,\ldots,b_n\right) g\left(a_1,a_2,\ldots,a_n\right)$, then you can only conclude that $f = \lambda g$ and $g = \lambda^{-1} f$ for some scalar factor $\lambda$. You can try to scratch out the $\lambda$ by substituting things for $a_i$ and $b_i$, but it isn't easy to come up with good things to substitute, since everything will vanish once any two $a_i$ (or any two $b_i$) are equal. ... $\endgroup$ – darij grinberg Nov 1 at 16:40
  • $\begingroup$ ... I like your idea, but I think it's missing some parts to be a complete proof. $\endgroup$ – darij grinberg Nov 1 at 16:41
  • $\begingroup$ Oh, I recall it can be proved by the number of roots. Since $deg(f(b_1,⋯,b_n)\le (n-1)^2$, it can not have $n(n-1)$ roots. Since each $a_k^j$ ($1\le k \le n, 1\le j \le n-1$) is a root of $R_1-R_2=0$, it vanishes and so we get the identity. $\endgroup$ – Math Wizard Nov 1 at 17:05
  • $\begingroup$ $R_1 - R_2 = 0$ is clear. What's not clear is the value of the $\lambda$ factor. $\endgroup$ – darij grinberg Nov 1 at 17:10
0
$\begingroup$

Consider sequence $A067689$ in the website oeis.org . This sequence consists of the reciprocals of the functions of $n$ that you seek. These achieve astronomical values very quickly. For example, for $n=9$, the function yields a value of just over 50 quattourdecillion. Why not be satisfied with entering the first few values in a table and referring to them thus rather than inflicting upon yourself the pain of calculating them?

$\endgroup$

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.