In connection to previous posts here and herenow we know $2\cos\frac{2\pi}{5}$ can be represented as $\sqrt{2-\sqrt{2+...}}$ simply represented as $cin\sqrt2[1-1+]$ and this converges to inverse of golden ratio $\frac{1}{\phi}$here
Let us examine the finite radicals
$\sqrt2 = 2\cos\frac{1}{4}\pi$
$\sqrt{2-\sqrt2} = 2\cos\frac{3}{2^3}\pi$
$\sqrt{2-\sqrt{2+\sqrt2}}$ ==> $n\sqrt2(1-1+)$ = $2\cos\frac{7}{2^4}\pi$
$n\sqrt2(1-1+1-)$ = $2\cos\frac{13}{2^5}\pi$
$n\sqrt2(1-1+1-1+)$ = $2\cos\frac{25}{2^6}\pi$
$n\sqrt2(1-1+1-1+1-)$ = $2\cos\frac{51}{2^7}\pi$
$n\sqrt2(1-1+1-1+1-1-)$ = $2\cos\frac{103}{2^8}\pi$
Let us analyse the angle pattern observed in Geometric progression approach
$\frac{3}{2^3}=\frac{1}{2^2}+\frac{1}{2^3}$
$\frac{7}{2^4}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}$
$\frac{13}{2^5}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}- \frac{1}{2^5}$
$\frac{25}{2^5}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}- \frac{1}{2^5}- \frac{1}{2^6}$
If we explore further we can observe a pattern as $++--$ of $2^{-n}$ after $\frac{1}{4}$
In this series again interesting pattern of getting numerator is as follows
$\frac{2^3}{5}$ = $1+\frac{3}{5}$ ( $1$ is the numerator)
$\frac{2^4}{5}$ = $3+\frac{1}{5}$ ( $3$ is the numerator)
$\frac{2^5}{5}$ = $6+\frac{2}{5}$ ( $6+1 = 7$ is the numerator)
$\frac{2^6}{5}$ = $12+\frac{4}{5}$ ( $12+1 = 13$ is the numerator)
$\frac{2^7}{5}$ = $25+\frac{3}{5}$ ( $25$ is the numerator)
$\frac{2^8}{5}$ = $51+\frac{1}{5}$ ( $51$ is the numerator)
$\frac{2^9}{5}$ = $102+\frac{2}{5}$ ( $102+1 = 103$ is the numerator) and so on
Summing infinite Geometric series clearly results in $\frac{2}{5}$ which is amazing and $cin\sqrt2[1-1+] = 2\cos\frac{2}{5}\pi$
Astonishingly modulo of 5 with respect to $2^n$ exactly predicts the sign for the component of the geometric series
$2^3 \equiv 3 \pmod 5 $
$2^4 \equiv 1 \pmod 5 $
$2^5 \equiv 2 \pmod 5 $
$2^6 \equiv 4 \pmod 5 $
Subsequently it repeats. Here important thing to observe is odd number modulo translates to $+$ sign and even number modulo translates to $-$ sign in Geometric series.
Therefore just by taking modulo for $2^n$ of any odd number(like 5 in this case) we can predict the pattern of geometric series with respect to $2^{-n}$ in deriving the result of cyclic infinite nested square roots of 2
It is exciting to know that the numerator sequence is associated with integer sequence of Expansion of $\frac{(1-x+2x^2)}{((1-2x)(1+x^2))}$ this type. But I don't know this kind of expansion how it is done to get integer sequences. Please explain
Please help me to express the infinite Geometric series in some better way in summation notation. (Even previous one also)
