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

sponsoredwalk
Messages
531
Reaction score
5
\begin{bmatrix}<br /> 1 &amp; \frac{1}{2} &amp; \frac{1}{3} \ldots &amp; \ldots &amp; \frac{1}{n}\\<br /> <br /> \frac{1}{2} &amp; \frac{1}{3} &amp; \frac{1}{4} \ldots &amp; \ldots &amp; \frac{1}{n + 1}\\<br /> <br /> \frac{1}{3} &amp; \frac{1}{4} &amp; \right \frac{1}{5}\ldots &amp; \ldots &amp; \frac{1}{n + 2}\\<br /> <br /> \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\<br /> <br /> \frac{1}{n} &amp; \frac{1}{n + 1} &amp; \frac{1}{n + 2} &amp; \ldots &amp; \frac{1}{2n - 1}<br /> <br /> \end{bmatrix}


Express the individual entries h_i_jin 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?
 
Physics news on Phys.org
Well, you saw that every entry was a fraction with "1" in the numerator, didn't you?:-p

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.
 
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...
Back
Top