Intermediate Math Problem of the Week 12/11/2017

[PF PotW Robot]

Here is this week's intermediate math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods. Spoiler tags are optional.

Let $a_0 = \dfrac52$ and $a_k = a_{k-1}^2 - 2$ for $k \geq 1$. Compute $\displaystyle\prod_{k=0}^\infty \left(1 - \frac{1}{a_k} \right)$ in closed form.

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[Delta2]

Ok I am not attempting to give a solution just saying two facts that may inspire other people to find solutions
Wolfram solves the recurrence relation and gives $a_n=2^{2^n}+2^{-2^{n}}$ (I am too rusty on solving this and it is a non linear recurrence which makes it harder)
As for the product wolfram says it converges rapidly to about 0.428571
http://www.wolframalpha.com/input/?i=product+from+0+to+infinity+of+(1-1/(2^(-2^n)+2^(2^n)))

 

[fresh_42]

Ok I am not attempting to give a solution just saying two facts that may inspire other people to find solutions
Wolfram solves the recurrence relation and gives $a_n=2^{2^n}+2^{-2^{n}}$ (I am too rusty on solving this and it is a non linear recurrence which makes it harder)
As for the product wolfram says it converges rapidly to about 0.428571
http://www.wolframalpha.com/input/?i=product+from+0+to+infinity+of+(1-1/(2^(-2^n)+2^(2^n)))

Yes. This looks like $\dfrac{3}{7}=\dfrac{2^2-1}{2^3-1}$ which doesn't make it easier. $a_n=2^{2^n}+2^{-{2^n}}$ is easy to prove: $a_{n+1}=(2^{2^n}+2^{-2^{n}})^2 -2=2^{2\cdot 2^{n}}+2 \cdot 2^{2^{n}} \cdot 2^{-2^{n}} + 2^{-2 \cdot 2^{n}} -2 = 2^{2\cdot 2^{n}} + 2^{-2 \cdot 2^{n}} = 2^{2^{n+1}}+2^{-2^{n+1}}$ but this was only the fun part.

 

[Delta2]

I thought a bit of this problem I ended up for an expression for the product that involves $\lim\limits_{n->\infty}\frac{a_{n+1}+1}{\prod\limits_{k=0}^{n}{a_k}}$. Any ideas how to handle this term?

 

[Delta2]

Though probably one needs not to solve for the recurrence equation in order to calculate the product, here is how it can be done using standard method:

assume $a_n=x_n+\frac{1}{x_n}$ ,$x_n>0$ then plugging this in the recurrence relation we get $x_n+\frac{1}{x_n}=x^2_{n-1}+\frac{1}{x^2_{n-1}} (1)$. We immediately can notice that if $x_n=x^2_{n-1} (2)$ then (1) is automatically satisfied, so by repeatedly applying (2) we get $x_n=x_0^{2^n}$

and by $a_0=5/2=x_0+\frac{1}{x_0}$ we get $x_0=2, x_0=\frac{1}{2}$ so $x_n=2^{2^n}$ or $x_n=2^{-2^n}$ but both solutions leading to the same form for $a_n=x_n+\frac{1}{x_n}=2^{2^n}+2^{-2^n}$.

 

[StoneTemplePython]

The problem seems to have sigmoid function written all over it. If you look at the closed form of $2^{2^n}+2^{-2^n}$ it is reminiscent of the function for

$2\cosh(z) = e^{z} + e^{-z}$, and plugging in $e^x = z$ in there (which, taking advantage of positivity, can be done), except the problem uses base $2$ not $e$ for exponentiation. And if you plug $x_n + \frac{1}{x_n}$ into $\Big(1 - \frac{1}{a_k}\Big)$ you get $\frac{(x^2 - x + 1)}{(x^2 + 1)}$ for each term. The graph of this looks eerily familiar to a curious pade approximation of the hyperbolic tangent function I did a while back. In particular

$\frac{\sinh(x)}{\cosh(x)} = \tanh(x) \approx \frac{x}{\frac{x^{2}}{\frac{x^{2}}{\frac{x^{2}}{\frac{x^{2}}{\frac{x^{2}}{11} + 9} + 7} + 5} + 3} + 1} = \frac{21 x \left(x^{4} + 60 y^{2} + 495\right)}{x^{6} + 210 x^{4} + 4725 x^{2} + 10395}$

plotting the graphs over say $[-100, 100]$ and the similarities jump out. Note $f(x) = \frac{(x^2 - x + 1)}{(x^2 + 1)}$ is rescaled, 'flipped' (think determinant of -1) and shifted. You can also see the squaring and increment by two here. (our problem is nesting a recurrence that squares and decrements by 2... but even so.)

I think the solution is outside my reach, but the structure bugs me.

 

[fresh_42]

What's really mean is, if you know the answer is $\dfrac{3}{7}$ and your calculation gets you $\dfrac{4}{7}$.
And once again, a search for a missing sign and / or faulty index shift ...

 

[Delta2]

I get $\frac{2}{7}\lim\limits_{n->\infty}\frac{a_{n+1}+1}{\prod\limits_{k=0}^{n}{a_k}}$. That limit seems to converges to 3/2 however I just can't find a way to prove it. Can easily see that an upper bound for that limit is 2, however the limit is not 2 its 3/2 :(.

 

[fresh_42]

I have the same, just that I thought it is two, when it has to be 3/2.