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

Posted 9 years ago
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 Does anyone know the answer to this:The sum of sqrt(x^2-p^2) from x= p+1 to neither exactly or approximately?In particular when n=(p^2+1)/2thankyou, there is an interesting application for it.J.
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Posted 9 years ago
 Thanks, If anyone is interested it is to do with this problem. Quite simple to describe, but not so easy to do.Imagine we draw an infinite number of circles of integer values 1,2,3... radius, centred on each of these points on the x axis (0,0), (1,0), (2,0)...i) How many of these circles cross the y axis between 0 and p (not including 0 or p)? Is there a formula in terms of p.A computer program needed the sum requested or could evaluate it in a loop, which is slowing it down.Let's consider p=10For radius 1 there is one crossing point (not including the origin), from (0,0) at (0,1). For radius 2 there are two from (0,0) and (1,0)...up to radius 10, 10 points -1 =9 (as 0,10) doesn't count. For radius 11 there are 6 from (0,5)...(0,10), the first few are two high, (0,11) etc..don't reach.To get the number 6 the program did 11-1-int(sqrt(11^2-10^2) and a loop is needed to evaluate the number of crossing points for radius 12,13...up to int((10^+1)/2) = 50, as sqrt(51^2-50^2) >10 i.e. for radius 51, or more, no circle crosses between 0 and 10.for p=10^n, the results were n=1 124. n=2 23460, n=3 3,493,008 n=4 468,382,827 n=5 currently being evaluated. It seems 0.5p^2ln(p) is quite a good match and the formula above, seems to lead to something similar. The int function may account for the discrepancy..?, 3/4 of the square root numbers are crossing points but not sqrt(2), sqrt(6), sqrt(10)...ii) If, instead of the y axis we use x=d...(0
Posted 9 years ago
 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.