2

For $a_1=\sqrt a , a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt {a}}} ,\ldots $ , We have $a_n$ as $$\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1$$ where $A=\sqrt{4a-7}$

Note: We have $- , + , +$ every three terms.

How we can prove this relation? (I want to prove it in general not only $a=2$ and period is $3$ not $4$)

Frank
  • 5,984
S.H.W
  • 4,379
  • In this page , there is no answer for it – S.H.W Nov 06 '16 at 12:45
  • And in this page , uses "Wolfy" and I don't know what is that mean. – S.H.W Nov 06 '16 at 12:54
  • Essentially, you have the form$$\sqrt{a\pm\sqrt{a\pm\sqrt{a\pm\sqrt{a\pm\ldots}}}}$$Setting $x$ at the right spot will give you a polynomial. With one of its roots as the convergence. For the theorem listed above, the operators cyclic by $3$. Namely $-,+,+$. So you'll be finding the roots of an octic ($2^3$). For period $4$, you'll be finding the roots of a $16$ degree polynomial, and so on. – Frank Nov 06 '16 at 18:37

1 Answers1

1

Your limit is a root of the degree $8$ polynomial $(((x^2-a)^2-a)^2-a)-x$.

As seen in Exploring 3-cycle points for quadratic iterations, if $b = \frac {1+\sqrt{4a-7}}2$

it factors over $\Bbb Q(b)$ as $(x^2-a-x)P(x)\overline{P}(x)$

where $P(x) = x^3+bx^2+(b-a-1)x+(a-ab-1)$
and $\overline P(x) = x^3+\overline bx^2+(\overline b-a-1)x+(a-a\overline b-1) \\ = x^3+(1-b)x^2+(-a-b)x+(ab-1))$.

The root you are interested in is not a fixpoint of $x \mapsto x^2-a$ so it is a root of one of the two cubics. Since the signs are $+,+,-$, the cubic should have two positive and one negative root when $a \ge 7/4$

When $a \to \infty, a \sim b^2$, so the second cubic is "equivalent" to $x^3-bx^2-b^2x+b^3 = (x-b)^2(x+b)$, so this is the right one, and you want to solve it.

Now, if $y=x^2-a$ and $z=y^2-a$ are the two conjugates of your root $x$ then Galois theory predicts that $(x+\zeta_3y+\zeta_3^2z)^3$ is a polynomial expression in $b$ (and $\zeta_3$), so we need to find it, take the right cube root, then deduce $x,y,z$ from $x+\zeta_3y+\zeta_3^2z$, its conjugate, and $x+y+z$.

$(x+\zeta_3y+\zeta_3^2z)^3 - (x+y+z)^3= 3((\zeta_3-1)(x^2y+y^2z+z^2x)+(\zeta_3^2-1)(xy^2+yz^2+zx^2))$

But
$x^2y+y^2z+z^2x = y(y+a)+z(z+a)+x(x+a) \\ = x^2+y^2+z^2+a(x+y+z) \\ = (x+y+z)(a+x+y+z) - 2(xy+yz+zx) = (b-1)(a+b-1)-2(-a-b)\\ = ab+b^2+a+1 \\ = ab+(a+b-2)+a+1 \\ = ab+2a+b-1 $

and
$xy^2+yz^2+zx^2 = x(z+a)+y(x+a)+z(y+a) \\ = xz+yx+zy+a(x+y+z) \\ = (-a-b)+a(b-1) \\ = ab-2a-b $

so that
$(x+\zeta_3y+\zeta_3^2z)^3 \\ = (b-1)^3 - \frac 92 (2ab+1) + \frac 32 \sqrt{-3}(4a+2b-1) \\ = (ab-2a-b+3) - \frac 92 (2ab+1) + \frac 32 \sqrt{-3}(4a+2b-1) \\ = -(8ab+2a+b)+\frac{15}2 + \frac 32 \sqrt{-3}(4a+2b-1) \\ = ((-2b+\frac 12) + \frac 32 \sqrt{-3})(4a+2b-1) \\ = -(4a+2b-1)^{\frac 32} \exp(i \arctan {\frac {3\sqrt 3}{1-4b}}) $

Since $x,z >0$ and $y<0$, the argument of $(x+\zeta_3y+\zeta_3^2z)$ is the one near $-\pi/3$, and so

$(x+\zeta_3y+\zeta_3^2z) = (4a+2b-1)^{\frac 12} \exp(i \frac 13( \arctan {\frac {3\sqrt 3}{1-4b} - \pi}))$

Taking twice the real part, we obtain

$2x-y-z = 2(4a+2b-1)^{\frac 12} \cos(\frac 13( \arctan {\frac {3\sqrt 3}{1-4b} - \pi}))$

Adding $x+y+z = b-1$, we obtain

$3x = b-1 + 2(4a+2b-1)^{\frac 12} \cos(\frac 13( \arctan {\frac {3\sqrt 3}{1-4b}} - \pi))$

since $\cos(\frac 13( \arctan {\frac {3\sqrt 3}{1-4b}} - \pi)) = \cos(\frac 13( \arctan {\frac {3\sqrt 3}{4b-1}} + \pi)) = \cos(\frac 13( \frac \pi 2 - \arctan {\frac {4b-1}{3\sqrt 3}} + \pi)) \\ = \cos(\frac \pi 2 - \frac 13( \arctan {\frac {4b-1}{3\sqrt 3}})) = \sin( \frac 13( \arctan {\frac {4b-1}{3\sqrt 3}}))$

we obtain the same formula.

Community
  • 1
mercio
  • 50,180
  • It seems too good but I can't understand it! Can you send some pages to me for better understanding ? – S.H.W Nov 06 '16 at 17:23
  • have you ever seen how to solve cubic equations ? – mercio Nov 06 '16 at 17:54
  • Yes , but in that cases cubic equations decomposes to one degree and quadratic components and it was easy to solve . But when I search in the internet for solving cubic equation, I find cardano's method. But I can't use it for this question – S.H.W Nov 06 '16 at 18:58
  • well my answer is cardano's solution in disguise though. Maybe there's a smarter way to get the formula because things simplified a bit miraculously – mercio Nov 06 '16 at 19:00
  • Do you know any good sources for the solving cubic equation step by step ? – S.H.W Nov 06 '16 at 19:02
  • anything that explain cardano's method. – mercio Nov 06 '16 at 19:03
  • Also I don't know meaning of zeta in your solution. – S.H.W Nov 06 '16 at 19:07
  • $\zeta_3 = \frac {-1+ \sqrt {-3}}2$ is a primitive cube root of $1$ – mercio Nov 06 '16 at 19:52