Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nxn Hilbert Matrix terms

  1. Feb 21, 2010 #1
    [tex] \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} [/tex]


    Express the individual entries [tex]h_i_j[/tex]in terms of i & j.


    The answer is

    [tex]h_i_j = \frac{1}{i + j - 1} [/tex]

    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?
     
  2. jcsd
  3. Feb 22, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Well, you saw that every entry was a fraction with "1" in the numerator, didn't you?:tongue2:

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook