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 $n=4 \text{(even case)} \hspace{3cm} D_4=\frac1{}-\frac1{}-\frac1{}-\frac1{}\frac1{}-\frac1{}-\frac1{}-\frac1{}-\frac1{}-\frac1{}-\frac1{}-\frac1{}-\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}+\frac1{}$.

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

$D_5=\frac1{}- {^5C_2} \text{terms of the form} \frac1{x_1x_2x_3x_4x_5}, \text{where any three terms say} x_1,x_2,x_3$ 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?

  • $\begingroup$ 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

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

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

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 $$ |A|=\dfrac{\prod\limits_{1\le i<j\le n} (j-i)^2}{\prod\limits_{i,j=1}^{n}(i+j)} $$

  • $\begingroup$ 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

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?


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