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Ranger-of-trente-deux-glands
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$$\int_{0}^{\pi}Si(x)Si(x)dx$$

Problem:
Let: $$P=\frac{2}{3^{2}}+\frac{5}{2\cdot7^{2}}+\frac{11}{5\cdot13^{2}}+\frac{17}{11\cdot19^{2}}+\frac{23}{17\cdot29^{2}}+\frac{31}{23\cdot37^{2}}+\cdots+\frac{157}{149\cdot163^{2}}+\frac{167}{157\cdot173^{2}}$$.

$$P=\frac{2}{3^{2}}+\sum_{n=1}^{37}\frac{p_{n+2}}{p_np^2_{n+3}}$$, where $$p_n$$ is the n th prime number.

Then show: $$\left|\frac{\ln\left(P\right)}{\pi^{2}}\right|>0.12345678910\cdots$$.

A Conjecture :

There exists $$a_i,c_i$$ strictly positive reals and $$b_i,d_i$$ positive integers such that $$x\in(0,\alpha)$$ ,$$\exists\alpha<1/2$$ then we have :

$$\frac{\arctan(2x)}{2x}=-1+\frac{-1+\sqrt{1+4x^2}}{2(-1+\sqrt{1+x^2})}-a_0f(x)-\sum_{n=1}^{\infty}a_n(f(x^{\frac{1}{n}}))^{2n}-\sum_{i=1}^{\infty}c_ix^{b_i}(1-x)^{d_i},f(x)=\frac{\sinh(x)}{x}-1$$

Deleted question by pillars :

Recently I discover a simple algorithm (like Heron) using fixed-point to have "any real roots of some particular polynomials i.e The Fibonacci's polynomials" with some constraint.

Algorithm:

Let $$f(x)=\sum_{i=0}^{n}a_ix^i$$. Then we want :

$$f(x_{root})=0$$

Now we introduce a second polynomial where

$$b_{k}=\left(-1\right)^{k}a_{k}^{2}+2\sum_{j=0}^{k-1}\left(-1\right)^{j}a_{j}a_{2k-j}$$

The second polynomial is (take the numerator):

$$x\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)-C=0\tag{I}$$

Or:

$$x\left(b_{0}\left(x+1\right)+b_{1}\cdot\frac{1}{x+1}+b_{2}\frac{x+1}{x+2}+b_{3}\cdot\frac{x+2}{2x+3}+\cdot\cdot+\frac{b_{n}\left(F_{n-2}x+F_{n-1}\right)}{F_{n-1}x+F_{n}}\right)-C=0$$

Where $$F_n$$ are the Fibonacci's numbers

Now we isolate the roots:

$$x=\frac{C}{\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)}=g(x), g(x)\neq\pm \infty,x\geq 0,|g'(0)|<1,C>0$$

Now : $$g(g(g(\cdots (0)\cdots)))=x_{roots}$$

For a generalization see https://en.wikipedia.org/wiki/Continuant_(mathematics) where Fibonacci's number's appears too .

Question/Context :

Using the describe iteration above and Newton's method have we a degenerate case where it doesn't converges?

https://en.m.wikipedia.org/wiki/Fixed-point_subring

https://en.m.wikipedia.org/wiki/Deformation_ring

Example :

$$8((x+1)^3/8+x^3/8-1)=2x^3+3x^2+3x-7=(2x+2)(x(x+1/2+9/2×1/(x+1))-7/16)=0$$

It's take from last fermat theorem showing the roots is irrational .In the case $$n=5$$ see https://en.m.wikipedia.org/wiki/Quintic_function

For the case $$n=3$$ see my deleted answer :

Let $$x\geq 0$$ and define :

$$f(x)=\frac{2}{x+1/2+\frac{1}{x+1}},f(x)=x$$

The roots is equal to one

If so using Hm-Gm in the simplest way and taking the limit at infinity it's equal to one .If so provided it converges we have a general criterion for cubic .

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