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Recurrencies and summing by parts

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Edited by Valentin Fadeev, Thursday, 22 July 2010, 12:23

This question was posed in sci.math.research group on Google:

http://groups.google.com/group/sci.math.research/browse_thread/thread/48ca096ec5f08a39?hl=en

After latexing the mind-bending plain text of the discussion it looks like this:

a sub one equals one

equation left hand side a sub n equals right hand side one divided by one minus b super n times n ary summation from m equals one to n minus one over a sub m times matrix row 1 n row 2 m times open one minus b close super n minus m times b super m of asterisk operator

assuming b not equals one . The author's conjecture was that for large n :

a sub n b minus one divided by natural log of b

Summing by parts is quite a standard device. Though, like with integration by parts the difficult part often is use it at the right moment. Whittakker and Watson ascribe it's systematic introduction to Abel. Probably the best account of it is given in "Concrete Mathematics". The authors introduce "definite sums" which are effectively the sums with an omitted last term:

equation left hand side n ary summation from a to b over f of x times delta times x equals right hand side n ary summation from x equals a to x equals b minus one over f of x

The cryptic delta is added solely to enhance the analogy with definite integrals.

The general formula is the following:

equation left hand side n ary summation from n equals a to n equals b minus one over u times normal cap delta times v equals right hand side u times v vertical line sub n equals a super n equals b minus n ary summation from n equals a to n equals b minus one over cap e times v times normal cap delta times u

Where equation left hand side normal cap delta times f of x equals right hand side f times open x plus one close minus f of x is the difference operator and equation left hand side cap e times f of x equals right hand side f times open x plus one close is the shift operator. The formula is easily proved by evaluating normal cap delta times u times v

equation left hand side n ary summation from a to b over normal cap delta times f of x equals right hand side f of b minus f of a

Part of the sum (*) on the right prompts for the binomial formula. Hence, it would be good to pull it out of the sum. Let's try:

equation sequence u equals a sub m times normal cap delta times v equals matrix row 1 n row 2 m times open one minus b close super n minus m times b super m

equation left hand side normal cap delta times u equals right hand side a sub m plus one minus a sub m

equation sequence v equals n ary summation from m equals one to x over matrix row 1 n row 2 m times open one minus b close super n minus m times b super m vertical line sub one super n equals one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b left parenthesis one minus b times right parenthesis super n minus one

For

equation left hand side n ary summation from m equals one to n over matrix row 1 n row 2 m times open one minus b close super n minus m times b super m equals right hand side

equation left hand side equals right hand side n ary summation from m equals zero to n over open matrix row 1 n row 2 m times open one minus b close super n minus m times b super m close postfix minus left parenthesis equation left hand side one minus b times right parenthesis super n equals right hand side left parenthesis one minus b plus b times right parenthesis super n postfix minus left parenthesis equation left hand side one minus b times right parenthesis super n equals right hand side one postfix minus left parenthesis one minus b times right parenthesis super n

Putting it all together:

equation left hand side n ary summation from m equals one to n minus one over a sub m times matrix row 1 n row 2 m times open one minus b close super n minus m times b super m equals right hand side

equation left hand side equals right hand side a sub n left parenthesis one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b times open one minus b times right parenthesis super n minus one close minus n ary summation from m equals one to n minus one over open a sub m plus one minus a sub m close left parenthesis one postfix minus left parenthesis equation left hand side one minus b times right parenthesis super n minus n times b times open one minus b times right parenthesis super n minus one close equals right hand side

equation left hand side equals right hand side a sub n left parenthesis one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b times open one minus b times right parenthesis super n minus one close postfix minus left parenthesis one postfix minus left parenthesis equation left hand side one minus b times right parenthesis super n minus n times b times open one minus b times right parenthesis super n minus one close times open a sub n minus one close equals right hand side one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b left parenthesis one minus b times right parenthesis super n minus one

This gives an equation for a sub n :

equation left hand side a sub n times open one minus b super n close equals right hand side one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b left parenthesis one minus b times right parenthesis super n minus one

equation left hand side a sub n equals right hand side one postfix minus left parenthesis one minus b times right parenthesis super n minus n times b left parenthesis one minus b times right parenthesis super n minus one divided by one minus b super n

Permalink 2 comments (latest comment by Valentin Fadeev, Sunday, 25 July 2010, 21:50)
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