What is the Formula for Entries in an Nxn Hilbert Matrix?

[sponsoredwalk]

$$\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \ldots & \ldots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \ldots & \ldots & \frac{1}{n + 1}\\ \frac{1}{3} & \frac{1}{4} & \right \frac{1}{5}\ldots & \ldots & \frac{1}{n + 2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \frac{1}{n} & \frac{1}{n + 1} & \frac{1}{n + 2} & \ldots & \frac{1}{2n - 1} \end{bmatrix}$$


Express the individual entries $$h_i_j$$in terms of i & j.


The answer is

$$h_i_j = \frac{1}{i + j - 1}$$

but I can't for the life of me understand how you would recognize that this formula fits the pattern formed in the matrix.

If you were answering this question, would you just attempt to form a formula with n's, i's and j's floating around or is there some method that would be helpful when addressing these types of problems?

 

[HallsofIvy]

Well, you saw that every entry was a fraction with "1" in the numerator, didn't you?

So, it's just a matter of looking at the denominators.

In the first row, where i= 1, the denominators are: j=1, 1; j= 2, 2; j=3, 3; j= 4, 4; ... which are j= j+i-1.

In the second row, where i= 2, the denominators are: j= 1, 2; j= 2, 3; j= 3, 4; j= 4, 5;... which are j+ 1= j+ i- 1 again.