The 35 years ago

Algebra Level 4

1981 + 1981 1981 + 1981 1981 + = ? \sqrt{1981+\sqrt{1981-\sqrt{1981+\sqrt{1981-\sqrt{1981+\cdots}}}}}=\ ?


The answer is 45.

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1 solution

Qin Haichen
Apr 21, 2016

Let be A. If we square A. We get 1981 + . However, is also A. Hence A^2 = 1981 + (1981-A)^(1/2), A=45.

I keep asking this question: How do you know that it converges?

Otto Bretscher - 5 years, 1 month ago

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And its highly probable that you will not get a answer here too :-)

Rishabh Jain - 5 years, 1 month ago

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Why don't you defy the odds?

Or, better yet, prove the convergence of this one . Good luck with that! ;)

Otto Bretscher - 5 years, 1 month ago

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@Otto Bretscher Oh... Need to study for that.... Convergence is something which I came to know on briliant and I really need to work on that concept real hard... And still this question can be approached by:

x = 1981 + 1981 1981 + 1981 = 1981 + y x=\sqrt{1981+\sqrt{1981-\sqrt{1981+\sqrt{1981-\cdots}}}}=\sqrt{1981+y} y = 1981 1981 + 1981 1981 + = 1981 x y=\sqrt{1981-\sqrt{1981+\sqrt{1981-\sqrt{1981+\cdots}}}}=\sqrt{1981-x} Then squaring , manipulating and subtracting will do the job but I knew I would end up having convergence issues.. :-)

Rishabh Jain - 5 years, 1 month ago

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@Rishabh Jain We can consider the iteration function f ( x ) = 1981 + 1981 x f(x)=\sqrt{1981+\sqrt{1981-x}} with the seed a 0 = 1981 44.5 a_0=\sqrt{1981}\approx 44.5 . This is a very strong contraction, with f ( x ) < 0.001 |f'(x)|<0.001 on [ 44 , 46 ] [44,46] , so that the nested radical will converge to 45 very quickly.

Otto Bretscher - 5 years, 1 month ago

@Rishabh Jain The best way!!!

Abhi Kumbale - 5 years, 1 month ago

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