The ∞ \infty -radical madness

Algebra Level 5

Let R 1 = 2 + 2 + 2 − 2 − 2 − . . . R_1=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-...}}}}} be a 5-periodic nested radical, with respective sign pattern is ( + + − − − ) (++---) .

Let R 2 , R 3 , R 4 , R 5 R_2, R_3, R_4, R_5 be analogous radicals with sign patterns ( − + + − − ) (-++--) , ( − − + + − ) (--++-) , ( − − − + + ) (---++) , ( + − − − + ) (+---+) .

Find R 1 − R 5 + R 2 − R 4 + R 3 R_1-R_5+R_2-R_4+R_3 .

The problem is original.


The answer is 1.

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Max Filippov by Max Filippov , 26, Russia

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3 solutions

Max Filippov
Feb 3, 2016

We are going to use the formula 2 cos ⁡ x = 2 + 2 cos ⁡ 2 x 2\cos{x} = \sqrt{2+2\cos{2x}} which is valid for cos ⁡ x ≥ 0 \cos{x} \geq 0 .

2 cos ⁡ π 11 = 2 + 2 cos ⁡ 2 π 11 = 2 + 2 + 2 cos ⁡ 4 π 11 = 2 + 2 + 2 + 2 cos ⁡ 8 π 11 = 2\cos{\frac{\pi}{11}}= \sqrt{2+2\cos{\frac{2\pi}{11}}}= \sqrt{2+\sqrt{2+2\cos{\frac{4\pi}{11}}}} =\sqrt{2+\sqrt{2+\sqrt{2+2\cos{\frac{8\pi}{11}}}}}= = 2 + 2 + 2 − 2 cos ⁡ 3 π 11 = 2 + 2 + 2 − 2 + 2 cos ⁡ 6 π 11 = 2 + 2 + 2 − 2 − 2 cos ⁡ 5 π 11 = =\sqrt{2+\sqrt{2+\sqrt{2-2\cos{\frac{3\pi}{11}}}}}=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+2\cos{\frac{6\pi}{11}}}}}}=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos{\frac{5\pi}{11}}}}}}= = 2 + 2 + 2 − 2 − 2 + 2 cos ⁡ 10 π 11 = 2 + 2 + 2 − 2 − 2 − 2 cos ⁡ π 11 . =\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+2\cos{\frac{10\pi}{11}}}}}}}=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2\cos{\frac{\pi}{11}}}}}}}. It means precisely that R 1 = 2 cos ⁡ π 11 . R_1=2\cos{\frac{\pi}{11}}. After that, it is clear from the definitions of R i R_i that R 5 = R 1 2 − 2 = 2 cos ⁡ 2 π 11 R_5=R_1^2-2=2\cos{\frac{2\pi}{11}} , R 4 = R 5 2 − 2 = 2 cos ⁡ 4 π 11 R_4=R_5^2-2=2\cos{\frac{4\pi}{11}} , R 3 = 2 − R 4 2 = − 2 cos ⁡ 8 π 11 = 2 cos ⁡ 3 π 11 R_3=2-R_4^2=-2\cos{\frac{8\pi}{11}}=2\cos{\frac{3\pi}{11}} , R 2 = 2 − R 3 2 = − 2 cos ⁡ 6 π 11 = 2 cos ⁡ 5 π 11 . R_2=2-R_3^2=-2\cos{\frac{6\pi}{11}}=2\cos{\frac{5\pi}{11}}.

Thus, R 1 − R 5 + R 2 − R 4 + R 3 = 2 cos ⁡ π 11 − 2 cos ⁡ 2 π 11 + 2 cos ⁡ 3 π 11 − 2 cos ⁡ 4 π 11 + 2 cos ⁡ 5 π 11 = R_1-R_5+R_2-R_4+R_3=2\cos{\frac{\pi}{11}}-2\cos{\frac{2\pi}{11}}+2\cos{\frac{3\pi}{11}}-2\cos{\frac{4\pi}{11}}+2\cos{\frac{5\pi}{11}}= = 2 cos ⁡ π 11 + 2 cos ⁡ 9 π 11 + 2 cos ⁡ 3 π 11 + 2 cos ⁡ 7 π 11 + 2 cos ⁡ 5 π 11 = 2 × 1 2 = 1 . =2\cos{\frac{\pi}{11}}+2\cos{\frac{9\pi}{11}}+2\cos{\frac{3\pi}{11}}+2\cos{\frac{7\pi}{11}}+2\cos{\frac{5\pi}{11}}=2\times \frac{1}{2}=\boxed{1}.

4 Helpful 0 Interesting 0 Brilliant 2 Confused

With infinite series, you have to evaluate it as the limit of the truncated sequence.

Currently, you are assuming that a unique finite limit exists, which needs to be justified.

Calvin Lin Staff - 5 years, 4 months ago

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I might be wrong but since each R i R_i is less than 2 + 2 + 2 + . . . = 2 \sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2 , a finite limit exists.

shaurya gupta - 5 years, 4 months ago

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That only applies if all of the terms are positive. For example, 2 − 4 + 2 … \sqrt{ 2 - \sqrt{ 4 + \sqrt{ 2 \ldots }}} does not exist since we will have 2 − ( 2 + ϵ ) \sqrt{ 2 - ( 2 + \epsilon) } .

Besides, that needs to be stated in the solution in order for the argument to be justified (which is the point in my comment).

Calvin Lin Staff - 5 years, 4 months ago

Here is a specific example where the "initial iteration setup" leads to the wrong answer.

Especially in cases where we can solve for 2 (or more) finite limiting values, there is uncertainty about which value is the true limit. In certain other cases, there might be finite limiting values, but the sequence either doesn't converge, or tends off to infinity.

Calvin Lin Staff - 5 years, 4 months ago
Sahil Goyat
Nov 8, 2019 
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J=10000                                                            
l=[]
for j in range (J+1,J+6):
    s=2
    for i in range(0,j,5):
        s=2+((-1)**(((((i%5)+1)%4)%3)%2))*(s**0.5)
    l=l+[s]
print(l[0]-l[4]+l[1]-l[3]+l[2])                               #then you get the output 1.0

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Is there any other easy method or formula??

0 Helpful 0 Interesting 0 Brilliant 2 Confused

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