LibriumArcana.Games

The feared topic arrives!

I was messing around on mathematica, and apparently

Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1] = sqrt2 - 1.

Is there a way to prove this?

Surlethe wrote:I was messing around on mathematica, and apparently Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1] = sqrt2 - 1. Is there a way to prove this?

I assume that the notation Int[f,a,b] means integral of f from a to b. Let Ω = Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1]; then Ω = Int[x,Ω,1] = 1/2 - Ω²/2. The negative fixed point is unstable, while the positive is stable. QED.

Kuroneko wrote:

Surlethe wrote:I was messing around on mathematica, and apparently Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1] = sqrt2 - 1. Is there a way to prove this?

I assume that the notation Int[f,a,b] means integral of f from a to b. Let Ω = Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1]; then Ω = Int[x,Ω,1] = 1/2 - Ω²/2. The negative fixed point is unstable, while the positive is stable. QED.

:king:

I second Cleric. Kuroneko, are you like a math PHD? *awe*

I'm gonna try and wrap my head around Surlys math.

Kuroneko wrote:

Surlethe wrote:I was messing around on mathematica, and apparently Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1] = sqrt2 - 1. Is there a way to prove this?

I assume that the notation Int[f,a,b] means integral of f from a to b.

Yes. That was my intent.

Let Ω = Int[x, Int[x, Int[x, Int[x, ..., 1], 1], 1], 1]; then Ω = Int[x,Ω,1] = 1/2 - Ω²/2. The negative fixed point is unstable, while the positive is stable. QED.

OK. Is it possible to solve for a general continuous, differentiable function f(x) in the integrand? This morning, a friend and I figured out the solution you just gave; and then we were wondering about the general case of Ω = Int[f(x),Ω,n]. We got it to Ω' = f(n) - f(Ω). Is there a way to find an explicit formula for Ω?