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 know the following recurrence relation

$$a_n=\frac{a+na_{n-1}}{a-n}$$

with $a_0=1$ can be represented alternatively as an integral

$$a_n=a\int_0^1{x^{a-n-1}(2-x)^ndx}$$

Verifying this is easy, but is there any general technique to do this kind of transformations?

share|improve this question
    
Do you have a specific "class" or recurrence relations in mind? Otherwise the answer might be "No". –  Aryabhata Mar 5 '12 at 0:49
    
@Aryabhata: What conclusions can you draw in general? Even for $$a_n=a\int_0^1{x^{a-n-1}(b-cx)^ndx}$$? –  littleEinstein Mar 5 '12 at 0:55
    
I don't understand. Are you looking for general techniques which take you from a recurrence relation to integral, or are you asking if getting an integral representation is any useful? –  Aryabhata Mar 5 '12 at 0:58
add comment

1 Answer 1

If a polynomial formula for the $a_n$ term is known, then the numerator of each term can be multipled by $x^{y-1}$ where $y$ is the denominator of that term.

$$ \textstyle a_n=\frac{a}{a-n}+\frac{ n(n-1)a}{(a-n)(a-(n-1))}+\frac{n(n-1)(n-2)a}{(a-n)(a-(n-1))(a-(n-2))}+ ... +\frac{n!(a+a_0)(a-n)!}{(a-1)!} $$

$$\begin{eqnarray} a_n&=&\int_0^1 ( ax^{a-n-1}+nax^{(a-n)(a-(n-1))-1}\\ &&+n(n-1)ax^{(a-n)(a-(n-1))(a-(n-2))-1}+\dots\\ &&+n!(a+a_0)x^{\frac{(a-1)!}{(a-n)!}-1})dx\\ &=&\int_0^1 a_0 n!x^{\frac{(a-1)!}{(a-n)!}-1} + \sum_{k=0}^n k! {n\choose k} x^{k! {a-1\choose k}-1 } \text dx \end{eqnarray} $$

This is not the simplest integral form, but at least it allows $a_n$ to be calculated when $n$ is not an integer.

share|improve this answer
add comment

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.