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This is my code so far:

For[i=2, i<10, i++, Solve[x^i == x+1 && x >0]]

I would like to sum up all the x values that it outputs. How can I do so?

Total@Table[x /. First@NSolve[x^i == x + 1 && x > 0], {i, 2, 9}] — MarcoB, Dec 23 '19 at 21:20
Obligatory reference: Why should I avoid the For loop in Mathematica? — MarcoB, Dec 23 '19 at 21:22

score 5 · Answer 1 · answered Dec 23 '19 at 21:23

5

Quick answer

Total@Table[ x /. Solve[x^i == x + 1 && x > 0][[1]] , {i, 2, 9}]//N
(*9.76035*)

answered Dec 23 '19 at 21:23

Ulrich Neumann

I ran the program by the way. It computes all the numbers positive numbers x s.t. x^n = x+1. Numbers like 1.618, 1.3247, 1.2207, etc. Its a sequence approaching 1 in the limit. Then I added all the decimal parts: .618 + .3247 + .2207 etc and found that it was divergent. Then I took the first 1000 sums and compared it to the first 10000 sums and I found a constant ratio of 1.596. Is there a concept for this? The ratio of divergent sums computed at logarithmic intervals. – Philip Yeranosian Dec 24 '19 at 02:43
1

+1, also: Sum[x /. Solve[x^i == x + 1 && x > 0][[1]], {i, 2, 9}] // N – WReach Dec 24 '19 at 06:38

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