Solving for terms of the sequence given by $ x^{2(2^k-1)}(x-1)=\ln2$ or gain some insight

Question

I'm interested in properties of the sequence given by (int k) :

$$ x^{2(2^k-1)}(x-1)= \ln2$$

$k = 1$: $x^2(x-1) = \ln2$ or $x^3-x^2 = \ln2$ :$x \approx 1.3737...$

$k=2$: $x^6(x - 1) = \ln2$ or $x^7 - x^6 = \ln2$ : $x\approx1.2175...$

$k=3$: $x^{14}(x-1) = \ln2$ or $x^{15} - x^{14} = \ln2$: $x\approx1.1284...$

Obviously this is asking for the base when certain consecutive powers differ by $\ln2$. Also apparent is how it approaches 1. Unfortunately, I'm not able to solve them generally. Might there be some recursive property from one element to the next. Or maybe some series representation.

Answer 1

Since $P_k(x)=x^{2(2^k-1)}(x-1)-\ln 2$ is a polynomial whose leading term is an odd power, there is at least a real solution to $P(x)=0$. That's because $\lim_{x\rightarrow-\infty} P(x)=-\infty$ and $\lim_{x\rightarrow+\infty} P(x)=+\infty$ and $P$ is continuous.

Next, note that if $P(x_k)=0$, and we now know that such an $x_k$ exists, then you can't expect a closed form for it. That's because there is no such formula for polynomials of degree above 4. It's the Abel-Rufini theorem.

Still, let's show that $x_k\rightarrow 1$.

First, notice that $$x_k^{2(2^k-1)}(x_k-1)=\ln 2$$ implies that $x_k>1$. Indeed, since the right-hand side is positive, so is the left-hand side. And since $2(2^{k-1}-1)$ is even, $x_k^{2(2^k-1)}>0$. So $x_k>1$.

We can now write $x_k=1+h_k$ with $h_k>0$. Now $$(1+h_{k+1})^{2(2^{k+1}-1)}h_{k+1}=\ln 2=(1+h_{k})^{2(2^{k}-1)}h_{k}$$ It's quite easy to see that it's necessary for $h_{k+1}\leq h_k$ (left as an exercise). So the sequence $\{h_k\}$ decreases and is positive. So it must be converge to a limit $h\geq 0$.

Letting $h_k$ converge to $h$ in $$\ln 2=(1+h_{k})^{2(2^{k}-1)}h_{k}\tag{1}$$ implies that $h=0$. So this proves that $x_k\rightarrow 1$.

Answer 2

One can make approximations of the solution building, around $x=1$, the $[1,n]$ Padé approximant of $$x^{2(2^k-1)}(x-1)- \log(2)$$

To fit in the page, I limited to $n=4$ and $a=\log(2)$. The approximation is $$x=1+\frac{4 a^4 \left(2^k-2\right) \left(2^k-1\right) \left(2^{k+1}-3\right)+12 a^3 \left(-9\ 2^k+4^{k+1}+5\right)+36 a^2 \left(2^k-1\right)+6 a}{a \left(2^k-1\right) \left(a \left(a \left(2^{k+1}-3\right) \left(a \left(2^k-2\right) \left(2^{k+1}-5\right)+2^{k+5}-40\right)+18 \left(3\ 2^{k+1}-7\right)\right)+48\right)+6}$$ This would give the following results $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 1.36953 & 1.36955 \\ 2 & 1.21408 & 1.21522 \\ 3 & 1.12998 & 1.12815 \\ 4 & 1.07960 & 1.07632 \\ 5 & 1.04724 & 1.04507 \\ 6 & 1.02664 & 1.02630 \\ 7 & 1.01434 & 1.01516 \\ 8 & 1.00747 & 1.00864 \\ 9 & 1.00382 & 1.00486 \\ 10 & 1.00193 & 1.00271 \end{array} \right)$$ and I suppose that using this extimate as $x_0$, Newton method should converge in a couple of iterations.