MHB Show that it satisfies the equation

[mathmari]

Hey!

How can I show that $2 \cos{\left ( \frac{2 \pi }{5} \right )}$ satisfies the equation $x^2+x-1=0$ ?? (Wondering)

 

[chisigma]

Hey!

How can I show that $2 \cos{\left ( \frac{2 \pi }{5} \right )}$ satisfies the equation $x^2+x-1=0$ ?? (Wondering)

Starting from the trigonometric identity...

$\displaystyle cos 5\ \theta = 16\ \cos^{5} \theta - 20\ \cos^{3} \theta + 5\ \cos \theta\ (1)$

... setting $x = \cos \theta$ the equation $\cos 5\ \theta= 1$ becomes...

$\displaystyle 16\ x^{5} - 20\ x^{3} + 5\ x - 1 = (x-1)\ (4\ x^{2} + 2\ x - 1)^{2} = 0\ (2)$

... and the soltions of (2) are $\cos 0 = 1$, $\cos \frac{2\ \pi}{5} = \frac{- 1 + \sqrt{5}}{4}$ and $\cos \frac{4\ \pi}{5} = \frac{-1 - \sqrt{5}}{4}$. The rest is easy and is left to You...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

Hey!

How can I show that $2 \cos{\left ( \frac{2 \pi }{5} \right )}$ satisfies the equation $x^2+x-1=0$ ?? (Wondering)

Hi! (Blush)

Substitute $2\cos\frac {2\pi}{5} = e^{2\pi i/5}+e^{-2\pi i/5}$? (Wondering)

 

[mathmari]

Starting from the trigonometric identity...

$\displaystyle cos 5\ \theta = 16\ \cos^{5} \theta - 20\ \cos^{3} \theta + 5\ \cos \theta\ (1)$

... setting $x = \cos \theta$ the equation $\cos 5\ \theta= 1$ becomes...

Why do we set $\cos 5\ \theta= 1$ ?? (Wondering)

 

[mathmari]

Substitute $2\cos\frac {2\pi}{5} = e^{2\pi i/5}+e^{-2\pi i/5}$? (Wondering)

$$\left (2 \cos{\left (\frac{2 \pi}{5}\right )}\right)^2+2 \cos{\left (\frac{2 \pi}{5}\right )}+1=\left ( e^{2\pi i/5}+e^{-2\pi i/5}\right )^2+
e^{2\pi i/5}+e^{-2\pi i/5}+1 \\ =e^{4\pi i/5}+2+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}+1=e^{4\pi i/5}+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}+3$$

How could I continue?? (Wondering)

 

[chisigma]

Why do we set $\cos 5\ \theta= 1$ ?? (Wondering)

... because $\cos (5\ \frac{2\ \pi}{5})= \cos (2\ \pi) = 1$...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

$$\left (2 \cos{\left (\frac{2 \pi}{5}\right )}\right)^2+2 \cos{\left (\frac{2 \pi}{5}\right )}+1=\left ( e^{2\pi i/5}+e^{-2\pi i/5}\right )^2+
e^{2\pi i/5}+e^{-2\pi i/5}+1 \\ =e^{4\pi i/5}+2+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}+1=e^{4\pi i/5}+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}+3$$

How could I continue?? (Wondering)

This is a geometric sequence and:
$$\sum_{k=0}^n a^{}r^k = a\frac{1-r^{n+1}}{1-r}$$

Oh. And you changed a $-$ into a $+$.

 

[chisigma]

It can be of some interest to know that an explicit expression derived from quadratic equations [ie expression containing square roots ...] for the amount $\displaystyle \cos \frac{2\ \pi}{n}$ exists if n is a Fermat prime, ie a prime for which is $\displaystyle n= 2^{2^{k}} + 1$. So far we only know the first five Fermat $F_{0}=3$,$F_{1} = 5$, $F_{2}=17$, $F_{3}=257$ and $F_{4}= 65537$. By what follows that the construction by non graded ruler and compass of a regular polygon of n sides is possible if n is a Fermat prime. For n = 3 and n = 5, the procedure has been found by Euclid. For n = 17 the procedure has been found in the nineteenth century. For n = 257 I think it was recently found as yet has not been found for n = 65537 ...

... that's quite a homework for a young MHB talent!;)...

Kind regards

$\chi$ $\sigma$

 

[mathmari]

Starting from the trigonometric identity...

$\displaystyle cos 5\ \theta = 16\ \cos^{5} \theta - 20\ \cos^{3} \theta + 5\ \cos \theta\ (1)$

... setting $x = \cos \theta$ the equation $\cos 5\ \theta= 1$ becomes...

$\displaystyle 16\ x^{5} - 20\ x^{3} + 5\ x - 1 = (x-1)\ (4\ x^{2} + 2\ x - 1)^{2} = 0\ (2)$

... and the soltions of (2) are $\cos 0 = 1$, $\cos \frac{2\ \pi}{5} = \frac{- 1 + \sqrt{5}}{4}$ and $\cos \frac{4\ \pi}{5} = \frac{-1 - \sqrt{5}}{4}$. The rest is easy and is left to You...

Kind regards

$\chi$ $\sigma$

$\displaystyle \cos 5\ \theta = 16\ \cos^{5} \theta - 20\ \cos^{3} \theta + 5\ \cos \theta\ $

$\theta=\frac{2 \pi}{5} : $

$\displaystyle \cos 5\ \frac{2 \pi}{5} = 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5}\ \\ \Rightarrow
1 = 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5}\ \\ \Rightarrow 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5} -1=0\ $

Setting $x=\cos \frac{2 \pi}{5}$ we have the following:

$$16\ x^5 - 20\ x^3 + 5\ x -1=0 \Rightarrow (x-1)\ (4\ x^{2} + 2\ x - 1)^{2} = 0$$

Since $x \neq 1$, $x$ must satisfy $(4\ x^{2} + 2\ x - 1)^{2}=0$, that means that $x$ has to satify the equation $4x^2+2x-1=0 \Rightarrow (2x)^2+(2x)-1=0$, which follows that $2x$ satisfies the equation $x^2+x-1=0$.

Therefore, we conlcude that $2 \cos \frac{2 \pi}{5} $ has to satisfy the equation $\ x^2 + \ x - 1=0$.

Is this correct?? (Wondering)

 

[mathmari]

This is a geometric sequence and:
$$\sum_{k=0}^n a^{}r^k = a\frac{1-r^{n+1}}{1-r}$$

$$\left (2 \cos{\left (\frac{2 \pi}{5}\right )}\right)^2+2 \cos{\left (\frac{2 \pi}{5}\right )}-1=\left ( e^{2\pi i/5}+e^{-2\pi i/5}\right )^2+
e^{2\pi i/5}+e^{-2\pi i/5}-1 \\ =e^{4\pi i/5}+2+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}-1=e^{4\pi i/5}+e^{-4\pi i/5}+e^{2\pi i/5}+e^{-2\pi i/5}+1$$

How is this a geometric sequence, when some powers are positive and other negative??

Do I have to write the terms with negative power in the form $\frac{1}{e^x}$ ?? (Wondering)

Oh. And you changed a $-$ into a $+$.

(Tmi)

 

[mathmari]

It can be of some interest to know that an explicit expression derived from quadratic equations [ie expression containing square roots ...] for the amount $\displaystyle \cos \frac{2\ \pi}{n}$ exists if n is a Fermat prime, ie a prime for which is $\displaystyle n= 2^{2^{k}} + 1$. So far we only know the first five Fermat $F_{0}=3$,$F_{1} = 5$, $F_{2}=17$, $F_{3}=257$ and $F_{4}= 65537$. By what follows that the construction by non graded ruler and compass of a regular polygon of n sides is possible if n is a Fermat prime. For n = 3 and n = 5, the procedure has been found by Euclid. For n = 17 the procedure has been found in the nineteenth century. For n = 257 I think it was recently found as yet has not been found for n = 65537 ...

... that's quite a homework for a young MHB talent!;)...

Kind regards

$\chi$ $\sigma$

Interesting! (Nerd)

 

[chisigma]

$\displaystyle \cos 5\ \theta = 16\ \cos^{5} \theta - 20\ \cos^{3} \theta + 5\ \cos \theta\ $

$\theta=\frac{2 \pi}{5} : $

$\displaystyle \cos 5\ \frac{2 \pi}{5} = 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5}\ \\ \Rightarrow
1 = 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5}\ \\ \Rightarrow 16\ \cos^{5} \frac{2 \pi}{5} - 20\ \cos^{3} \frac{2 \pi}{5} + 5\ \cos \frac{2 \pi}{5} -1=0\ $

Setting $x=\cos \frac{2 \pi}{5}$ we have the following:

$$16\ x^5 - 20\ x^3 + 5\ x -1=0 \Rightarrow (x-1)\ (4\ x^{2} + 2\ x - 1)^{2} = 0$$

Since $x \neq 1$, $x$ must satisfy $(4\ x^{2} + 2\ x - 1)^{2}=0$, that means that $x$ has to satify the equation $4x^2+2x-1=0 \Rightarrow (2x)^2+(2x)-1=0$, which follows that $2x$ satisfies the equation $x^2+x-1=0$.

Therefore, we conlcude that $2 \cos \frac{2 \pi}{5} $ has to satisfy the equation $\ x^2 + \ x - 1=0$.

Is this correct?? (Wondering)

Yes, it is!...

Kind regards

$\chi$ $\sigma$

 

[mathmari]

Yes, it is!...

Kind regards

$\chi$ $\sigma$

Great! Thank you very much! (Smile)