Nested Radical Equation

[Kharakov]

Hi, found an equation that converges to specific values, wondered if it has a name so I can read more about it. One of the values it converges to is Pi.

 

I'll eventually add in a table of generated values, in addition I'd like to note that I haven't been able to find the values online (the values the equation generates, except for Pi).

 

I've got some text based Maxima format equations if someone would like me to post them and run checks to verify the equation in Maxima (they should work in Mathematica as well, although you may need to tweak them a bit). If you do so, remember to set fpprec (floating point precision in Maxima, don't know what it's called in Mathematica) to at least 256 (perhaps 512 or higher), if you want the equations to work.

 

 

[imath]f(x,n,a) = \sqrt[n]{x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}}\ \times (n^{1/n}\ \times x^{\frac{n-1}{n}})^a [/imath]

 

a= number of iterations of radicals

 

For example, in the equation above a= 4 (if we disregard the ...).

 

I'm not positive, and do not have the literature to check it, but using the values (2,2,a approaches infinity) in the equation:

 

[imath]f(2.0,2.0,a)\to \pi \ as \ a \to \infty[/imath]

 

may correspond to one of Viete's formulas for Pi. (The equation does approach Pi for these values, I'm just unsure of whether it is one of Viete's formulas)

 

 

In addition (with a still representing number of iterations of radical):

 

[imath]g(x,n,a) = \sqrt[n]{x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}} [/imath]

 

[imath]z_1(x,n) = \lim_{a\to\infty} \ \frac{g(x,n,a)}{g(x,n,a+1)} \ = \ n^{1/n}\ \times x^{\frac{n-1}{n}}[/imath]

 

Which happens to be the only number that doesn't cause divergence when taken to the a power and multiplied times g(x,n,a) as a increases.

 

Notice that:

 

[imath]f(x,n,a)= g(x,n,a) \ \times \ [z_1(x,n)]^a[/imath]

 

 

Additional variations include (a-1 is the number of radical iterations in the following equation):

 

[imath]

h(x,n,a)=[f(x,n,a)]^n[/imath]

 

[imath]h(x,n,a)= {x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}}\ \times (n \ \times x^{n-1})^a [/imath]

 

a = 4 in the above equation (a - 1 = 3 so a = 4)

 

[imath]i(x,n,a)= {x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}}[/imath]

 

[imath]z_2(x,n) = \lim_{a\to\infty} \ \frac{i(x,n,a)}{i(x,n,a+1)} \ = \ n \times x^{n-1}[/imath]

 

This limit happens to be the derivative of [imath]x^n[/imath].

 

Notice that:

 

[imath]h(x,n,a)= i(x,n,a) \ \times \ [z_2(x,n)]^a[/imath]

 

Also notice that:

 

[imath]z_2(x,n) = [z_1(x,n)]^n[/imath]

 

 

Thanks for your time, Benesi