Ramanujan's nested radical is great! (2)

Algebra Level 5

2016 2016 2016 2016 2016 ⋯ 5 4 3 2016 2016 2016 2016 2016 ⋯ = 201 6 e + x \large \frac{2016\sqrt{2016\sqrt[3]{2016\sqrt[4]{2016\sqrt[5]{2016\cdots}}}}}{\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\cdots}}}}}}=2016^{e+x}

Find x x which satisfies the above equation.

Hint : Consider the infinite sum series of e e

Clarification : e ≈ 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is -2.

Tommy Li by Tommy Li , 22, Hong Kong

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3 solutions

Tommy Li
Jun 29, 2016

Relevant wiki: Nested Functions

e = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + … e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots

e = 1 + 1 + 1 2 ( 1 + 1 3 ( 1 + 1 4 ( 1 + 1 5 ( 1 + … ) ) ) ) e=1+1+\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+\frac{1}{5}(1+\dots))))

e − 1 = 1 + 1 2 ( 1 + 1 3 ( 1 + 1 4 ( 1 + 1 5 ( 1 + … ) ) ) ) e-1=1+\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+\frac{1}{5}(1+\dots))))

201 6 e − 1 = 201 6 1 + 1 2 ( 1 + 1 3 ( 1 + 1 4 ( 1 + 1 5 ( 1 + … ) ) ) ) 2016^{e-1}=2016^{1+\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+\frac{1}{5}(1+\dots))))}

201 6 e − 1 = 2016 2016 2016 2016 2016... 5 4 3 2016^{e-1}= 2016\sqrt{2016\sqrt[3]{2016\sqrt[4]{2016\sqrt[5]{2016...}}}}

2016 = 201 6 2 = 2016 2016 = 2016 2016 2016 2016 2016... 2016=\sqrt{2016^2}=\sqrt{2016\sqrt{2016}}=\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016...}}}}}

2016 2016 2016 2016 2016... 5 4 3 2016 2016 2016 2016 2016... = 201 6 e − 1 − 1 \large \frac{2016\sqrt{2016\sqrt[3]{2016\sqrt[4]{2016\sqrt[5]{2016...}}}}}{\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016...}}}}}}=2016^{e-1-1}

⇒ x = − 2 \Rightarrow x=-2

12 Helpful 0 Interesting 1 Brilliant 0 Confused

Taking logs (base 2016):

log ⁡ 2016 2016 2016 2016 ⋯ 4 3 = 1 + 1 2 ( 1 + 1 3 ( 1 + 1 4 ( 1 + ⋯ ) ) ) = 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + ⋯ = e − 1 ; \log 2016\sqrt{2016\sqrt[3]{2016\sqrt[4]{2016\cdots}}} = 1 + \tfrac12(1 + \tfrac13(1 + \tfrac14(1 + \cdots))) \\ = \frac1{1!} + \frac 1{2!} + \frac1{3!} + \frac1{4!} + \cdots = e - 1; log ⁡ 2016 2016 2016 2016 ⋯ = 1 2 ( 1 + 1 2 ( 1 + 1 2 ( 1 + 1 2 ( 1 + ⋯ ) ) ) ) = 1 2 1 + 1 2 2 + 1 2 3 + 1 2 4 + ⋯ = 1 ; \log \sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\cdots}}}} = \tfrac12(1 + \tfrac12(1 + \tfrac12(1 + \tfrac12(1 + \cdots)))) \\ = \frac1{2^1} + \frac1{2^2} + \frac1{2^3} + \frac1{2^4} + \cdots = 1; ∴ log ⁡ 2016 2016 2016 2016 ⋯ 4 3 2016 2016 2016 2016 ⋯ = ( e − 1 ) − 1 = e − 2 , \therefore \log \frac{2016\sqrt{2016\sqrt[3]{2016\sqrt[4]{2016\cdots}}}}{\sqrt{2016\sqrt{2016\sqrt{2016\sqrt{2016\cdots}}}}} = (e-1) - 1 = e-2, so that x = − 2 x = \boxed{-2} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Frank Petiprin
Feb 18, 2018 
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'''Basic Idea is to approximate the  value of the Ramanujan nested radicals of the 
left side of the equation with a numerical method and them use Logs to the base e
to solve for X
Log(NR1/NR2) = Log(2016**(X+e)) = (X+e)*Log(2016) 
NR1 approx 476481.481262981
NR2 approx 2015.999999999
Ratio = NR1/NR2=236.3499411026
Solve for X   =  Log(236.3499411026)/Log(2016) - e '''  
#                   BEGIN PYTHONISTA PROGRAM ON IPAD PRO**
import math
#A LOOP OF 52 GAVE THE BEST APPROXIMATIONS TO THE NESTED RADICALS 
beg = 1
end = 52
exp = end + 1
Dprodrt = ( 2016 )**.5
Nprodrt = ( 2016 )**(1/(end  + 1))
for loop in range(  beg, end + 1, 1  ):
    Dprodrt = ( 2016*Dprodrt )**.5
    exp -= 1
    Nprodrt = (2016*Nprodrt)**(1/exp)
print('NR1: Numerator', Nprodrt)
Ratio = Nprodrt/Dprodrt
print('NR2: Denomanator', Dprodrt)
print('Ratio ',Ratio,'Loop Count', end)
Logval = ( math.log(Ratio)/math.log(2016) ) - math.exp(1)
print('Answer X', Logval)
input('Ramanujan')
#               END PROGRAM

NR1: Numerator 476481.4812629815
NR2: Denomanator 2015.999999999
Ratio  236.3499411026 Loop Count 52
Answer X -2.0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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