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How can we prove the following equality?

$$\pi = \frac{10}{3}\;\sum_{n=0}^\infty\frac{(-1)^{n}}{2n+1}\left(\frac{\varphi^{2}}{3-\varphi}\right)^{n+ \frac{1}{2} }$$ where $\varphi=1.618\ldots$ is the golden ratio.

my work:

$$\sum _ {n=0} ^ \infty \frac{ (-1)^{n} }{2n+1} ( \frac{ \varphi ^{2} }{3- \varphi } )^{n+ \frac{1}{2} }$$ is really simillar to the taylor series for $\tan^{-1}(x)$ which is: $$\tan^{-1}(x)=\sum _ {n=0} ^ \infty \frac{ (-1)^{n} }{2n+1} (x )^{2n+1}$$ now let $x=( \frac{ \varphi ^{2} }{3- \varphi } )^{\frac{1}{2}}$ then we have:

$$x=( \frac{ \varphi ^{2} }{3- \varphi } )^{\frac{1}{2}} = ( \frac{ 1+\varphi }{3- \varphi } )^{\frac{1}{2}} = \sqrt{\frac{1+\frac{1+\sqrt{5}}{2}}{3-\frac{1+\sqrt{5}}{2}}} = \sqrt{\frac{\frac{3+\sqrt{5}}{2}}{\frac{5-\sqrt{5}}{2}}} = \sqrt{\frac{3+\sqrt{5}}{5-\sqrt{5}}} = \sqrt{\frac{20+8\sqrt{5}}{20}} = \sqrt{1+\frac{2\sqrt{5}}{5}}$$ now if $\pi=\frac{10}{3}\tan^{-1}(x)$ then $\tan^{-1}(x)=\frac{3\pi}{10}$ so $\tan(\frac{3\pi}{10})=x=\sqrt{1+\frac{2\sqrt{5}}{5}}$

then how can we show that $\tan(\frac{3\pi}{10})=\sqrt{1+\frac{2\sqrt{5}}{5}}$

Mathieu Rundström
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kn2798
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1 Answers1

5

Note that $\frac{3\pi}{10} +\frac{\pi}{5} = \frac{\pi}{2}$. This implies $\tan{\frac{3\pi}{10}} = \cot{\frac{\pi}{5}} = \frac{\cos{\frac{\pi}{5}}}{\sqrt{1-\cos^2\frac{\pi}{5}}}$. Hence, it is sufficient to find a value for $\cos \frac{\pi}{5}$, which we now do.

Now, as linked in the comments, there is a short derivation for $\cos{\frac{2\pi}{5}} = \frac{\sqrt{5}-1}{4}$. Let us proceed similarly to find $\cos{\frac{\pi}{5}}$. Let $x=\cos{\frac{\pi}{5}}=\frac{z+z^{-1}}{2}$, for $z=e^{i\pi/5}$. Since $\frac{\pi}{5}=\frac{2\pi}{10}$, the tenth cyclotomic polynomial thus equals zero at $z$, i.e., $z^4-z^3+z^2-z+1=0$. Since $z\neq 0$, we equivalently have $z^2-z+1-z^{-1}+z^{-2}=0$. Aiming to express this in terms of $x$, we have $z^2-z^{-2} = (z-z^{-1})^2-2 = 4x^2-2$, and $z-z^{-1}=2x$. Thus, $4x^2-2x-1=0,$ implying $x=\frac{1}{4}(1+ \sqrt 5)$, taking the positive root. Now what remains is just algebraic manipulations,

$$ \tan{\frac{3\pi}{10}} = \frac{\frac{1}{4}(1+ \sqrt 5)}{\sqrt{1-\left(\frac{1}{4}(1+ \sqrt 5)\right)^2}} = \frac{\frac{1}{4}(1+ \sqrt 5)}{\frac{1}{4}\sqrt{10-2\sqrt 5}} = \frac{1+ \sqrt 5}{\sqrt{10-2\sqrt 5}}. $$ Let us square this expression:

$$ \left(\frac{1+ \sqrt 5}{\sqrt{10-2\sqrt 5}}\right)^2 = \frac{6+2\sqrt 5}{10-2\sqrt 5} = \frac{3+\sqrt 5}{5-\sqrt 5}. $$ Rationalizing the denominator: $$ \frac{3+\sqrt 5}{5-\sqrt 5}\cdot \frac{5+\sqrt 5}{5+\sqrt 5} = \frac{20+8\sqrt 5}{20} = 1+\frac{2\sqrt 5}{5}. $$

Thus

$$ \tan{\frac{3\pi}{10}} = \sqrt{1+\frac{2\sqrt 5}{5}}, $$

as desired.

Mathieu Rundström
  • 3,647