Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently i have discovered a new infinite radical with non-constant (that means they are not the same and not periodic) coefficients. But i don't have a strict proof of this identity... The identity is

$$\sqrt[3]{23+\sqrt[3]{54+\sqrt[3]{972+\sqrt[3]{21870+\sqrt[3]{551124+\sqrt[3]{14526054+...}}}}}}=3.$$

The numbers are (besides 23) $3^{2n+1}(3^{n-1}+1)$. I am interested in how can we strictly prove that the limit is $3$.

share|improve this question
2  
Fine that you don't have the proof, yet you believe that nested radical equals $\;3\;$ : why? What have led you to believe so? –  DonAntonio Dec 31 '13 at 14:28
    
$3=\sqrt {1+2\cdot4}=\sqrt {1+2\cdot\sqrt {1+3\cdot5}}... $ This radical I got by similar manipulations, I know that they are informal, but I am sure in this case it is ok, and Wolfram Mathematica confirms it up to many decimal places. –  Elensil Dec 31 '13 at 14:31
    
And how does that relate to your radical? –  DonAntonio Dec 31 '13 at 14:32
3  
Hrm. It sounds like if you actually showed your thoughts, you could have spared me from doing pretty much all the work I put into my answer. –  Hurkyl Dec 31 '13 at 14:34
2  
Saying "new" is a radical move. There are more than a handful where "new" things were actually known decades, and sometimes centuries ago to people. –  Asaf Karagila Dec 31 '13 at 14:37
show 4 more comments

2 Answers

Let $a_0$ be your nested radical, and let $a_n$ be the radical that starts at the $n$-th term rather than at $23$.

They satisfy a recurrence relation:

$$ a_0^3 - 23 = a_1 $$ $$ a_m^3 - 3^{2m+1}(3^{m-1} + 1) = a_{m+1} $$

Assuming $a_0 = 3$, the first few terms would be

  • $a_1 = 4$
  • $a_2 = 10$
  • $a_3 = 28$
  • $a_4 = 82$

In fact, I conjecture $a_m = 3^m + 1$ if $m > 0$.

Let $b_m = (a_m - 1)/3^m - 1$. Then, $a_m = 3^m (1+b_m) + 1$ and the recurrence relation becomes

$$ (3^m (1+b_m) + 1)^3 - 3^{2m+1}(3^{m-1} + 1) = 3^{m+1} (1+b_{m+1}) + 1$$

which simplifies to

$$ (3^{2m-1} (b_m^2 + 3b_m + 3) + 3^{m} (b_m + 2) + 1) b_m = b_{m+1}$$

It's clear that if

$$ \lim_{x \to \infty} b_m $$

exists, then it has to equal zero. This confirms that $a_m = 3^m + 1$ for $m > 0$, and finally, that $a_0$ as conjectured.

This doesn't prove that the limit exists, though....

share|improve this answer
1  
If you are still interested in my thoughts : $3=\sqrt[3]{23+4}=\sqrt[3]{23+\sqrt[3]{54+10}}=\sqrt[3]{23+\sqrt[3]{54+\sqrt[3]{‌​972+28}}}$ Coefficients are exactly as i said because of the identity $(3^{n}+1)^3=3^{2n+1}(3^{n-1}+1)+3^{n+1}+1$ –  Elensil Dec 31 '13 at 14:59
add comment

Not quite sure what's there to ‘prove’, really. What you wrote basically boils down to:

$$3=\sqrt[3]{3^3}=\sqrt[3]{(3^3-4)+4}=\sqrt[3]{(3^3-4)+\sqrt[3]{4^3}}=\sqrt[3]{(3^3-4)+\sqrt[3]{(4^3-10)+10}}=\ldots,$$

where at each step the term being added and subtracted is of the form $a_{n+1}=3\cdot a_n-2$. Since $a_1$ $=4=3^1+1$, we have $a_n=3^n+1$, e.g., $a_2=10=9+1=3^2+1$. All that's left to show is that each step of the way, in order for the radical to make sense, the result of the subtraction is positive, i.e., $a_n^3-a_{n+1}>0\iff(3^n+1)^3-(3^{n+1}+1)>0\iff3^{3n}+3^{2n+1}>0$, which is indeed true for all real n.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.