vineri, 19 februarie 2021

AN EQUATION THAT, SURPRISING, IS NOT GRADE SEVEN ...

 In the book https://ogeometrie-cip.blogspot.com/2021/02/getting-to-know-new-problem-book.html

see PROBLEM 28, at page 115.


           I have not found a solution to this problem.

           I had to look at the solution given in the book. On page 120 it says just that...

(Long after that I found out about the author's solution: see below, in the contribution of Silviu Boga)

 
       I posted the problem on several forums, let's see what consequences it will have...

 Posted on Facebook, 2/19/2021 https://www.facebook.com/petre.ciobanu.5621/posts/708103799874909 

 
 Posted on AOPS, 2/20/2021 https://artofproblemsolving.com/community/c4h2458976_equation_for_sinpi14

 

 

SOLUTION CiP

Answer CiP 

 The roots of the polynomial are    $sin\frac{\pi}{14},\;\;-sin\frac{3\pi}{14}\;\;and\;\;sin\frac{5\pi}{14}$.

Or, complementarily expressed    $cos\frac{\pi}{7},\;\;-cos\frac{2\pi}{7}\;\;and\;\;cos\frac{3\pi}{7}$.

           

           Solution

                     

          According to de Moivre's formula

$(cos \alpha +\imath \cdot sin \alpha)^{7}=cos 7 \alpha+\imath \cdot sin 7\alpha,\;(cos \alpha-\imath \cdot sin \alpha)^{7}=cos 7 \alpha-\imath \cdot sin 7\alpha$

express $sin7\alpha=\frac{(cos\alpha+\imath \cdot sin \alpha)^{7}-(cos\alpha-\imath \cdot sin\alpha)^{7}}{2\imath}$. We develop $(cos \alpha \pm \imath \cdot sin\alpha)^{7}$ with Newton's binomial theorem , considering that $\imath ^{2}=\imath ^{6}=-1,\;\imath ^{3}=\imath ^{7}=-\imath,$

$\imath ^{4}=1,\;\imath ^{5}=\imath$, and we obtain  $\sin7\alpha=$

$=\frac{1}{2\imath}\cdot [cos^{7}\alpha+7\imath\; cos^{6}\alpha sin \alpha-21 cos^{5}\alpha sin^{2}\alpha-35\imath\; cos^{4}\alpha sin^{3} \alpha +35cos ^{3} \alpha sin^{4} \alpha +21 \imath\; cos^{2} \alpha sin^{5} \alpha -7 cos \alpha sin^{6} \alpha-$

$-\imath \; sin^{7} \alpha -(cos^{7}\alpha-7\imath\; cos^{6}\alpha sin \alpha -21 cos^{5}\alpha sin^{2}\alpha-35 \imath\; cos^{4}\alpha sin^{3}\alpha +35 cos^{3}\alpha sin^{4} \alpha -21 \imath \;cos^{2}\alpha sin^{5} \alpha -$

$-7 cos \alpha sin^{6} \alpha +\imath \; sin^{7} \alpha )]$. 

At the result    $sin 7\alpha=7cos^{6} sin \alpha -35 cos^{4} \alpha sin^{3} \alpha +21 cos^{2}\alpha sin^{5} \alpha -sin^{7} \alpha$

we replace further $cos^{2}\alpha=1-sin^{2}\alpha, \;\;cos^{4} \alpha =1-2sin^{2}\alpha +sin^{4}\alpha,\;$

$\;cos^{6}\alpha=1-3sin^{2}\alpha +3sin^{4}\alpha-sin^{6} \alpha$, and after other calculations we obtain 

(1)          $sin 7\alpha =-64sin^{7}\alpha+112sin^{5}\alpha-56sin^{3}\alpha +7sin \alpha$.

(See Bromwich's formulas (24) in Weisstein, Eric W. "Multiple-Angle Formulas.")

         We put in the formula (1) $\alpha=\frac{\pi}{14}$. Since $sin 7\cdot \frac{\pi}{14}=sin \frac{\pi}{2}=1$, we obtain that $x=sin \frac{\pi}{14}$ is the root of the seventh degree equation

(2)                     $ 64 \cdot x^{7}-112 \cdot x^{5}+56\cdot x^{3}-7\cdot x+1=0$.

           IF WE ARE LUCKY ENOUGH, we can factorize the polynomial on the left, and find the form equivalent to the equation (2)

$(x+1)\cdot (8\cdot x^{3}-4\cdot x^{2} -4 \cdot x+1)^{2}=0$.

 
          From the above line, it is clear to whom $x=sin\frac{\pi}{14}$ is the root.

$\blacksquare$ 

  ..................................................................................................................................

                REMARK 1

         We write formula (1) in the form

(3)          $sin7\alpha-1=-(sin\alpha+1)(8sin^{3}\alpha-4sin^{2}\alpha-4sin\alpha+1)^{2}$.

The values for which $sin7\alpha-1=0$ are $\alpha_{k}=\frac{\pi}{14}+k\cdot \frac{2\pi}{7},\;\;k \in \mathbb{Z}$. So

(4)          $(sin \alpha_{k}+1)(8sin^{3}\alpha_{k}-4sin^{2}\alpha_{k}-4sin \alpha_{k}+1)^{2}=0$.

        We analyze only cases $k\in \left \{0,\;1,...,6\;\right \}$ because $\alpha_{k+7m}\equiv \alpha_{k}\;\;(mod\;2\pi)$.

     The value $\alpha_{5}=\frac{\pi}{14}+\frac{10\pi}{7}=\frac{21\pi}{14}=\frac{3\pi}{2}$ corresponds to the case $sin\alpha_{5}=-1$, which makes zero the first factor in the  relation  (4). Further,

$sin\alpha_{0}=sin\alpha_{3}$, because $\alpha_{0}+\alpha_{3}=\frac{\pi}{14}+\frac{13\pi}{14}=\pi$,

$sin\alpha_{1}=sin\alpha_{2}$, because $\alpha_{1}+\alpha_{2}=\frac{5\pi}{14}+\frac{9\pi}{14}=\pi$,

$sin\alpha_{4}=sin\alpha_{6}$, because $\alpha_{4}+\alpha_{6}=\frac{17\pi}{14}+\frac{25\pi}{14}=3\pi$.

     So the given polynomial $8\cdot x^{3}-4\cdot x^{2}-4\cdot x+1$ has three different roots $sin\alpha_{0}=\sin\frac{\pi}{14},\;sin\alpha_{1}=sin\frac{5\pi}{14},\;sin\alpha_{4}=sin\frac{17\pi}{14}=sin(\pi+\frac{3\pi}{14})=-sin\frac{3\pi}{14}$. We get the first answer.

Observing that $\frac{\pi}{2}-\alpha_{0}=\frac{3\pi}{7},\;\frac{\pi}{2}-\alpha_{1}=\frac{\pi}{7},\;\frac{\pi}{2}-\frac{3\pi}{14}=\frac{2\pi}{7}$, we can still express the answer $x_{1}=cos\frac{\pi}{7},\;x_{2}=-cos\frac{2\pi}{7},\;x_{3}=cos\frac{3\pi}{7}$.

(End Rem #1)


                REMARK 2

           Writing, for the three roots found, some of Viete's relationships, we have identities

 (5)          $sin\frac{\pi}{14}-sin\frac{3\pi}{14}+sin\frac{5\pi}{14}=\frac{1}{2}$,

(6)          $cos\frac{\pi}{7}-cos\frac{2\pi}{7}+cos\frac{3\pi}{7}=\frac{1}{2}$,

(7)          $sin\frac{\pi}{14}\cdot sin\frac{3\pi}{14}\cdot sin\frac{5\pi}{14}=\frac{1}{8}$,

(8)          $cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7}\cdot cso\frac{3\pi}{7}=\frac{1}{8}$.

(End Rem #2)

$\bigstar$     $\bigstar$     $\bigstar$


 P.S.  Here ENDS the solitary adventure, starting from the attempt to solve the problem in the statement. I followed the path suggested by the indication in the book.With the price of a substantial number of scribbled sheets of paper, I am still glad that I solved more than was required:I determined ALL the roots of the given equation.

$\bigstar$      $\bigstar$      $\bigstar$

 

           During the days I worked on this problem, I was stimulated by the help of "fellow mathematicians" who sent me ideas or even other methods to solve this problem. I mention them below, trying not to miss any. My thanks to everyone.

 

  AOPS- the user vanstraelen suggested the development of $sin7\alpha$ in terms of $sin\alpha$.  He gets the formula (1) which he writes in the form (3).

(It all happened 65 minutes after I posted the question. Others here did not answe.)

 

Now, about those on Facebook.

Andreea Schier [19, Feb at 23:03] put up a photo


 I kind of twisted my neck, then thought about rotating it


     She chooses, as was most normal, to calculate the value of the polynomial $8x^{3}-4x^{2}-4x+1$ for $x=sin\frac{\pi}{14}$.

     It goes without saying that he uses the formula $8\cdot sin^{3}\alpha =6\cdot sin\alpha -2 \cdot sin3\alpha$ (obtained from $sin 3\alpha=3\cdot sin \alpha-4 \cdot sin^{3} \alpha$). However, it mentions the formula $sin^{2} \alpha=\frac{1-cos 2 \alpha}{2}$ as well as a transition from the cosine to the sine of the complementary angle. An expression is reached in the sinuses of various multiples of $\frac{\pi}{14}$, but only at the first power:

$S=2\cdot sin\frac{\pi}{14}-2\cdot sin\frac{3\pi}{14}+2\cdot sin \frac{5\pi}{14}-1$.

 She proves that $S=0$, calculating for this $S \cdot sin\frac{\pi}{7}$ ( it transforms sine products into sums with the formula $2\cdot sin \alpha \cdot sin \beta=cos(\alpha-\beta)-cos(\alpha+\beta)$ and all terms are canceled).

      I notice that, by the same calculation as hers, a demonstration of the Viete type relationship (5) is obtained.

 $\bigstar$

          In her comment  Marcelina Popa  starts from the equation $sin 4\alpha=sin 3\alpha$. This equation has in the open interval $(\;0,\;\pi)$ only the solutions $\alpha_{k}=(2k+1)\cdot \frac{\pi}{7}\;\;k=0,\;1,\;2$.


Equivalently, the equation is written successively (remember that in the interval $(\;0,\;\pi)$)

$4\cdot sin\alpha \cdot cos\alpha\cdot cos2\alpha=sin\alpha\cdot (3-4sin^{2}\alpha)$

$\overset{sin\alpha \neq 0}{\Leftrightarrow} 4\cdot cos \alpha\cdot (2cos^{2}\alpha-1)=3-4(1-cos^{2}\alpha)$

$\Leftrightarrow \;8\cdot cos^{3}\alpha-4\cdot cos^{2}\alpha-4\cdot cos \alpha+1=0$.

As a result, the equation  $8x^{3}-4x^{2}-4x+1=0$ is verified for $cos\frac{\pi}{7},\;cos\frac{3\pi}{7}=sin\frac{\pi}{14},\;cos\frac{5\pi}{7}$.

$\bigstar$


          Another calculation-based answer gave Dumitru Pietreanu. He starts from equality

$sin(\frac{3\pi}{14}+\frac{4\pi}{14})=sin\frac{\pi}{2}$,

which he develops using known formulas, finding that $x=sin\frac{\pi}{14}$ is the root of a polynomial $Q(x)$ and this can be expressed as $Q(x)=P^{2}(x)\cdot (x+1)$ where $P(x)$ is the given polynomial.

      Indeed, after the first step $sin\frac{3\pi}{14}\cdot cos\frac{4\pi}{14}+cos\frac{3\pi}{14}\cdot sin\frac{4\pi}{14}=1$. If $x=sin\frac{\pi}{14}$, then $cos^{2}\frac{\pi}{14}=1-x^{2},\;sin\frac{2\pi}{14}=2x\cdot cos\frac{\pi}{14},\;cos\frac{2\pi}{14}=1-2x^{2},$

$\;sin\frac{3\pi}{14}=3x-4x^{3},\;cos\frac{3\pi}{14}=cos\frac{\pi}{14}\cdot (4cos^{2}\frac{\pi}{14}-3)=cos\frac{\pi}{14}\cdot [4(1-x^{2})-3]=...,$

$sin\frac{4\pi}{14}=4x\cdot cos \frac{\pi}{14}\cdot cos\frac{2\pi}{14}=...,\;cos\frac{4\pi}{14}=2cos^{2}\frac{2\pi}{14}-1=...$.

In the end it results $-64x^{7}+112x^{5}-56x^{3}+7x=1$, so just the formula (2).

      He also congratulated Marcel Tena for his contribution.

 

$\bigstar$

 
          The one who found the mentioned book from other sources, also identified the solutions (two) of the author, on page 296, is

          Silviu Boga

 
          The first solution exploits the fact that $cos \frac{4\pi}{7}+\imath \cdot sin \frac{4\pi}{7} \neq 1$ verifies the equation $z^{7}=1$ (they are all $cos (k\cdot  \frac{2\pi}{7})+\imath \cdot sin (k\cdot\frac{2\pi}{7}), \;k=0,\;1,...,6$.
  !!! There was a small mistake, probably transcription, where I circled in the picture: it should be written   "u"  instead of  "1".  
          The second solution is similar to ours

$\bigstar$

 

          Ghimisi Dumitrel used the complex number $\zeta=cos\frac{\pi}{14}+\imath\cdot sin\frac{\pi}{14}$. We have $\zeta^{7}=\imath$. But $\zeta^{7}-\imath=\zeta^{7}+\imath^{7}=(\zeta+\imath)(\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1)$

$\Rightarrow \;\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1=0$.

 With $sin\frac{\pi}{14}=\frac{\zeta-\bar{\zeta}}{2}=\frac{\zeta^{2}-1}{2\imath z}$ is calculated $8sin^{3}\frac{\pi}{14}-4sin^{2}\frac{\pi}{14}-4sin\frac{\pi}{14}+1=$

$=8\cdot (\frac{\zeta^{2}-1}{2\imath z})^{3}-4\cdot (\frac{\zeta^{2}-1}{2\imath z})^{2}-4\cdot \frac{\zeta^{2}-1}{2\imath z}+1=-\frac{1}{\imath \zeta^{3}}\cdot [(\zeta^{2}-1)^{3}-\imath \zeta (\zeta^{2}-1)^{2}+2\zeta^{2}(\zeta^{2}-1)-\imath \zeta^{3}]=$

$=-\frac{1}{\imath \zeta^{3}}\cdot [\zeta^{6}-\imath \zeta^{5}-\zeta^{4}+\imath \zeta^{3}+\zeta^{2}-\imath \zeta-1]=0$.

 $\bigstar$


          Doru GHERASA  presented two methods of solving.

          In the first solution, he observes that $sin\frac{\pi}{14}=cos(\frac{\pi}{2}-\frac{\pi}{14})=cos\frac{6\pi}{14}$, so $\frac{\pi}{14}$ is a solution of the equation

(9)                     $sin \alpha =cos 6\alpha$.

 Equation (9) is successively transformed into $sin \alpha = 4cos^{3}2\alpha-3cos2\alpha $

$\Leftrightarrow\;sin\alpha=4(1-2sin^{2}\alpha)^{3}-3(1-2sin^{2}\alpha)\;\Leftrightarrow x=4(1-2x^{2})^{3}-3(1-2x^{2})$, so $sin\frac{\pi}{14}$ is a solution of equation

(10)                   $32x^{6}-48x^{4}+18x^{2}+x-1=0$.

He notices the factorization of (10) 

$(8x^{3}-4x^{2}-4x+1)(x+1)(4x^{2}-2x-1)=0$

 and, since $0<sin\frac{\pi}{14}<\frac{1}{2}$, it can be the root only for the first factor.

 

          In the second method, we start from  de Moivre's formula

$(a+\imath b)^{7}\overset{def}{=}(cos\frac{\pi}{14}+\imath sin \frac{\pi}{14})^{7}=cos\frac{\pi}{2}+\imath sin\frac{\pi}{2}=\;\imath$. 

The imaginary part of the equality, developing on the left side with Newton's binomial theorem involves

$7a^{6}b-35a^{4}b^{3}+21a^{2}b^{5}-b^{7}=1\;\;\;\overset{a^{2}=1-b^{2}}{\Rightarrow}\;\;\; 64b^{7}-112b^{5}+56b^{3}-7b+1=0$, similar to (2). Then mention that the last equation can be written

$(8b^{3}-4b^{2}-4b+1)^{2}(b+1)=0.$ 

 

$\bigstar$ 


          AT LAST, BUT NOT LEST, we thanks Mr Marcel ȚENA. I reported the problem to him for the first time. He also had that book. He is, among many other things, the author of the book "RADACINILE UNITATII".


In a discussion with him, I asked his opinion on which cyclotomic polynomial between $\Phi_{14}$ or $\Phi_{7}$ to use. From his answers I realized that it is possible with both.

          The solution now has only a few lines:

          The cyclotomic polynomial $\Phi_{14}$ has, by definition, one of its roots 

$\zeta=cos(3\cdot \frac{2\pi}{14})+\imath \cdot sin(3\cdot \frac{2\pi}{14})=cos\frac{3\pi}{7}+\imath \cdot sin\frac{3\pi}{7}$.

But $\Phi_{14}(X)=\Phi_{7}( -X)$ (Theorem 16, page 24) and 7 being prime number we have $\Phi_{7}(X)=X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1$ (Theorem 11, page 20). So $\zeta$ is a root of polynomial $X^{6}-X^{5}+X^{4}-X^{3}+X^{2}-X+1$.

      So $\zeta$ verifies the equation

$\zeta ^{6}-\zeta ^{5}+\zeta ^{4}-\zeta ^{3}+\zeta ^{2}-\zeta+1=0$

$\Leftrightarrow \;\;(\zeta ^{3}+\frac{1}{\zeta ^{3}})-(\zeta ^{2}+\frac{1}{\zeta ^{2}})+(\zeta+\frac{1}{\zeta})-1=0$.

 We replace above $\zeta +\frac{1}{\zeta}=2cos\frac{3\pi}{7}=sin(\frac{\pi}{2}-\frac{3\pi}{7})=sin \frac{\pi}{14}=2x$,

 $\zeta ^{2}+\frac{1}{\zeta ^{2}}=(\zeta +\frac{1}{\zeta})^{2}-2=(2x)^{2}-2=4x^{2}-2$,

 $\zeta ^{3}+\frac{1}{\zeta ^{3}}=(\zeta +\frac{1}{\zeta})^{3}-3(\zeta +\frac{1}{\zeta})=(2x)^{3}-3(2x)=8x^{3}-6x$

and we get the equation  $8x^{3}-4x^{2}-4x+1=0$.

$\blacksquare\;\blacksquare$ 

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