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sum of [sqrt(x^2 - p^2)] from x= p+1 to n

Posted 10 years ago

Does anyone know the answer to this:

The sum of sqrt(x^2-p^2) from x= p+1 to n

either exactly or approximately?

In particular when n=(p^2+1)/2

thankyou, there is an interesting application for it.

J.

POSTED BY: James Spencer
2 Replies

Approximating sum by integral

Integrate[(x^2-p^2)^1/2,{x,p + 1,(p^2+1)/2}]

This gives after simplification

1/8 (-1 + p^4 - 4 Sqrt[1 + 2 p] - 4 p Sqrt[1 + 2 p] + 
   4 p^2 (-2 Log[p] + Log[1 + p + Sqrt[1 + 2 p]]))

Better approximations can be obtained using Euler-Maclauren sum formula.

POSTED BY: S M Blinder
Posted 10 years ago
POSTED BY: James Spencer
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