solve for x if u can

Algebra Level 2

x = 1 + 2 1 + 3 1 + . . . . . x=\sqrt { 1+2\sqrt { 1+3\sqrt { 1+..... } } }


The answer is 3.

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1 solution

Abhi Abhimanyu
Jan 2, 2015

This can be solved by noting a more general formulation:

x= \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}} \, Setting this to F(x) and squaring both sides gives us:

F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}} \, Which can be simplified to:

F(x)^2 = ax+(n+a)^2 +xF(x+n) \, It can then be shown that:

F(x) = {x + n + a} \, So, setting a =0, n = 1, and x = 2:

3 = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \,

To type in latex, you just need to place it within the brackets \ ( Latex code \ ) (with no spaces between). I've updated the question for your reference.

Note that you have not provided a proof as yet that F ( x ) = x + n + a F(x) = x + n + a . All that you have shown is that x + n + a x + n + a is a possible solution to the recurrence relation that you created. There are many other solutions, like F ( x ) F(x) equal to infinity.

Calvin Lin Staff - 6 years, 5 months ago

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Can you email this method to me and some problems related to it

Sameer Marathe - 6 years, 1 month ago

This is also called as Ramanujan's Nested Radical.

Marc Vince Casimiro - 6 years, 5 months ago

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