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Feb 15 2018 09:55am
This is for calculating the risk of ruin on an infinite series of bets with multiple, constant possible outcomes ('x' is later used with a different equation).

So for example I have:

x = 0.546567*x^0.006 + 0.213617*x^1.006 + 0.129152*x^2.006 +0.074344*x^3.006 + 0.011277*x^4.006 + 0.011044*x^6.006 + 0.011502*x^9.006 + 0.002360*x^25.006 + 0.000111*x^49.606 + 0.000026*x^920.006

in which x ≈ 0.999635211420277

And I can also find the values of x as the last exponent gets larger using wolframalpha, for example:

x = 0.546567*x^0.006 + 0.213617*x^1.006 + 0.129152*x^2.006 +0.074344*x^3.006 + 0.011277*x^4.006 + 0.011044*x^6.006 + 0.011502*x^9.006 + 0.002360*x^25.006 + 0.000111*x^49.606 + 0.000026*x^1250
x ≈ 0.999269159069166

x = 0.546567*x^0.006 + 0.213617*x^1.006 + 0.129152*x^2.006 +0.074344*x^3.006 + 0.011277*x^4.006 + 0.011044*x^6.006 + 0.011502*x^9.006 + 0.002360*x^25.006 + 0.000111*x^49.606 + 0.000026*x^1300
x ≈ 0.999207863987952

x = 0.546567*x^0.006 + 0.213617*x^1.006 + 0.129152*x^2.006 +0.074344*x^3.006 + 0.011277*x^4.006 + 0.011044*x^6.006 + 0.011502*x^9.006 + 0.002360*x^25.006 + 0.000111*x^49.606 + 0.000026*x^1403
x ≈ 0.999142898697158

As the last exponent gets larger it seems x approaches somewhere around .999. I even plotted some points using the last exponent and x as coordinates to see an asymptotic curve form. But wolframalpha won't solve the equation anymore after the last exponent far exceeds 1400 as it just continually times out because it takes too much computational power.

I guess I'd like to know the limit of x as the last exponent approaches infinity, maybe using some other method besides the brute forcing as done here. The exact figure would be nice but I'd be happy with an approximation using an arbitrarily large last exponent too. This area of math is beyond me though.
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Feb 16 2018 11:47am
Let f(x) = 0.546567*x^0.006 + 0.213617*x^1.006 + 0.129152*x^2.006 +0.074344*x^3.006 + 0.011277*x^4.006 + 0.011044*x^6.006 + 0.011502*x^9.006 + 0.002360*x^25.006 + 0.000111*x^49.606 - x

(the fixed part of your equation)

f has 3 real roots :
α = 0
β ≈ 0.9988
γ ≈ 1.0112

You want to solve for x : f(x) + 2.6 10⁻⁵ xⁿ = 0 (1n)
α = 0 will remain a solution.
γ = 1 is a solution.

As n tends to + ∞,
2.6 10⁻⁵ xⁿ tends to zero for any real number x in [0,1[,
2.6 10⁻⁵ xⁿ tends to + ∞ for any real number x > 1.

Solutions for (1n) will be :
α = 0
βn → β ≈ 0.9988
γ = 1
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Feb 17 2018 09:37am
Thanks a bunch.
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