Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to find the simplest way to get an expression of the determinant of the following infinite matrix as m tends to infinity.

$$ \left[\begin{array}{cccccc} 1 & a_{1} & 0 & \cdots & 0 & 0 \\ \beta_{1} & 1 & a_{2} & \cdots & 0 & 0\\ 0 & \beta_{2} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & a_{m}\\ 0 & 0 & 0 & \cdots & \beta_{m} & 1 \end{array} \right]$$

I have considered using both the Leibniz formula or the Laplace formula. Leibniz formula required considering sums over permutations which I was hoping to avoid and Laplace formula seems somewhat recursive even though I have only 2-3 elements in each row it. Is there a simpler solution to this problem which I am overlooking ?

Edit: I am just after an algebraic expansion of the determinant into and infinite series of some kind ?

Any help would be much appreciated.

share|improve this question
    
The criteria for absolute convergence of this determinant is given by Whittaker & Watson section 2.82 p37 as an example due to von Koch. It is probably a good idea to confirm your case satisfies this necessary and sufficient condition. –  John Spitzer Sep 5 at 7:45

1 Answer 1

up vote 1 down vote accepted

Try for $\lambda=-1$

$$ \det(I-\lambda K) = \left[ \sum_{n=0}^\infty (-\lambda)^n \operatorname{Tr } K^n \right]= \exp{(\sum_{n=0}^\infty(-1)^{n+1}\frac{\operatorname{Tr} A^n}{n}z^n})$$

copied from

http://en.wikipedia.org/wiki/Fredholm_determinant

Note that $I-K$ can be put as $D^+ + D^-$, where $D^+$ are strict upper triangular and $D^-$ are strict lower diagonal. Then use the binomial theorem for $(D^+ + D^-)^n$.

Note that $(D^-)^k$ and $(D^+)^k$ are fairly simple to compute;)

share|improve this answer
    
I am not sure the method you are suggesting is much simpler, if anything it seems quite the opposite. –  Hardy Mar 6 '12 at 19:00
1  
I provided you with a further hint;) The binomial theorem is now doing the combinatorics for you;) –  plusepsilon.de Mar 7 '12 at 8:25
    
thanks i 'll try it out and would post back if i run into trouble. –  Hardy Mar 7 '12 at 9:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.