How to prove √5+√5+√5−√5+√5+√5+√5−⋯=2+√5+√15−6√52\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}

Ramanujan stated this radical in his lost notebook:

5+5+55+5+5+5=2+5+15652

I don’t have any idea on how to prove this.

Any help appreciated.

Thanks.

Answer

The correct period has length 4, namely (+,+,+,-)
x1=+5+5+55+5+5+5=2+5+15652=2.7472
The other roots of the quartic in x are given by the patterns (+,+,,+),(+,,+,+),(,+,+,+), respectively
x2=+5+55+5+5+55+=25+15+652=2.5473

x3=+55+5+5+55+5+=2+515652=1.4888

x4=5+5+5+55+5+5+=2515+652=2.7833

This immediately implies that the four roots obey the system,
x21=x2+5x22=x3+5x23=x4+5x24=x1+5
also studied by Ramanujan. (See this related post.) More generally, using any of the 24=16 possible periods,

x=±a±a±a±a±

will be the absolute value of a root of the 16th deg eqn,

x=(((x2a)2a)2a)2a

In his Notebooks IV (p.42-43), Ramanujan stated that (1) was a product of 4 quartic polynomials, one of which is the reducible,

(x2xa)(x2+xa+1)=0

and the other three had coefficients in the cubic,

y3+3y=4(1+ay)

Using Mathematica to factor (1), we find that it is indeed a product of (2) and a 12th deg eqn with coefficients in a. After some manipulation, the 12 roots are,

xn=yz4±12(y2)(y+z)z2y

where,

z=±y2+4

Since there are 4 sign changes and (3) gives 3 choices for y, this yields the 12 roots.

Note: For a=5 (as well as a=2), the cubic factors over Q, hence no cubic irrationalities are involved, and one of the xn will give the value of the appropriate infinite nested radical.

P.S. Interestingly, for period length n>4, not all the roots of the deg 2n equation will be expressible as finite radical expressions for general a. The exception is a=2 where the solution involves roots of unity as discussed in this post.

Attribution
Source : Link , Question Author : Shobhit , Answer Author : Community

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