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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?
Moderator: Charon
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.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?
Kuroneko wrote: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.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?
Yes. That was my intent.Kuroneko wrote:I assume that the notation Int[f,a,b] means integral of f from a to b.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?
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?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.