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Problem A. 546. (November 2011)

A. 546. Show that


\frac{1}{\sin^2{\dfrac{\pi}{4k+2}}} +
\frac{1}{\sin^2{\dfrac{3\pi}{4k+2}}} +
\frac{1}{\sin^2{\dfrac{5\pi}{4k+2}}} + \ldots +
\frac{1}{\sin^2{\dfrac{(2k-1)\pi}{4k+2}}} = 2k(k+1)

for every positive integer k.

(5 pont)

Deadline expired on December 12, 2011.


Solution 1. Let Un(x) be the nth Chebyshev-polynomial of the second kind, for which Un(cos t)sin t=sin ((n+1)t). The roots of U2k are the numbers x_1=\cos\frac{\pi}{2k+1}=\sin\frac{(2k-1)\pi}{4k+2}, x_2=\cos\frac{2\pi}{2k+1}=\sin\frac{(2k-3)\pi}{4k+2}, ..., x_k=\cos\frac{k\pi}{2k+1}=\sin\frac{\pi}{4k+2} and their negatives, xk+1=-x+1, xk+2=-x2, ..., x2k=-xk. On the left-hand side of the statement we have the sum of the squares of these numbers; the statement is equivalent with


\sum_{j=1}^{2k} \frac1{x_j^2} = 4k(k+1).  (1)

Lemma 1. U0(x)=1, U2(x)=4x2-1, and U2k+2(x)+U2k-2(x)=(4x2-2)U2k(x) for k\ge1.

Proof. From U0(cos t)sin t=sin t, we get U0(x)=1. Similarly, from U2(cos t)sin t=sin (3t)=(4cos2-1)sin t we get U2(x)=4x2-1.

From the identity sin (a+b)+sin (a-b)=2sin acos b we obtain

U2k+2(cos t)sin t+U2k-2(cos t)sin t=

=sin ((2k+3)t)+sin ((2k-1)t)=2sin ((2k+1)t)cos (2t)=

=2U2k(cos t)sin t.(2cos2-1),

so U2k+2(x)+U2k-2(x)=(4x2-2)U2k(x).

Lemma 2. Let U_{2k}(x)=a_{k,0}+a_{k,1}x+\ldots+a_{k,2k}x^{2k}. Then ak,0=(-1)k, ak,1=0 és ak,2=(-1)k+1.2k(k+1).

Proof. Apply induction on k. If k=0 then U0(x)=1, so a0,0=1=(-1)k, a0,1=0, and a0,2=0=-4k(k+1) If k=1 then U0(x)=4x2-1, so a1,0=-1=(-1)k, a1,1=0, and a1,2=4=2k(k+1). Hence, the lemma holds for k=0 and k=1.

For the induction step assume the lemma for k and (k-1). By Lemma 1,


a_{k+1,0}+a_{k+1,1}x+a_{k+1,2}x^2+\ldots=U_{2k+2}(x)=
(4x^2-2)U_{2k}(x)-U_{2k-2}(x)=

 =
(4x^2-2)(a_{k,0}+a_{k,1}x+a_{k,2}x^2+\ldots)
-(a_{k-1,0}+a_{k-1,1}x+a_{k-1,2}x^2+\ldots) =

 =
(4x^2-2)((-1)^k+(-1)^{k+1}\cdot 2k(k+1)x^2+\ldots)
-((-1)^{k-1}+(-1)^k\cdot2k(k-1)x^2+\ldots).

From this we read

ak+1,0=-2.(-1)k-(-1)k-1=(-1)k+1,

ak+1,1=0,

finally

ak+1,2=4.(-1)k-2.(-1)k+1.2k(k+1)-(-1)k.2k(k-1)=

=(-1)k(4+4k(k+1)-2k(k-1))=(-1)k+2.2(k+1)(k+2).

The lemma is proved.

Now from the Viéte formulae we find


\sum_{j=1}^{2k}\frac1{x_j^2} = 
\left(\sum_{j=1}^{2k}\frac1{x_j}\right)^2 - 2\sum_{1\le i<j\le 2k}
\frac1{x_ix_j} = 0^2 -2\frac{a_{k,2}}{a_{k,0}} =
 -2\frac{(-1)^{k+1}\cdot 2k(k+1)}{(-1)^k} = 4k(k+1).

Solution 2. Let \varepsilon=\cos\dfrac\pi{4k+2}+i\sin\dfrac\pi{4k+2}, then \ell egész számra \sin\dfrac{\ell\pi}{4k+2} =
\dfrac{\varepsilon^\ell-\varepsilon^{-\ell}}{2i}.

From the identity a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+b^{n-1}) we have


  \frac{1}{\sin{\frac{(2j-1)\pi}{4k+2}}}
  = \frac{2i}{\varepsilon^{2j-1}-\varepsilon^{-(2j-1)}} =


  = \frac{2i(\varepsilon^{2k(2j-1)}+\varepsilon^{(2k-2)(2j-1)}
    +\varepsilon^{(2k-4)(2j-1)}+\ldots+\varepsilon^{-2k(2j-1)})}{
    (\varepsilon^{2k(2j-1)}+\varepsilon^{(2k-2)(2j-1)}
    +\ldots+\varepsilon^{-2k(2j-1)})(\varepsilon^{2j-1}-\varepsilon^{-(2j-1)})}=


  = \frac{2i}{\varepsilon^{(2k+1)(2j-1)}-\varepsilon^{-(2k+1)(2j-1)}}
  \sum_{\ell=-k}^k \varepsilon^{2\ell(2j-1)}
  = \sum_{\ell=-k}^k \varepsilon^{2\ell(2j-1)},

so


\frac{1}{\sin^2{\frac{(2j-1)\pi}{4k+2}}}
= \left(\sum_{\ell=-k}^k \varepsilon^{2\ell(2j-1)}\right)^2
= (2k+1) + \sum_{\ell=1}^{2k} \bigl(2k+1-\ell)(\varepsilon^{2\ell(2j-1)}
+ \varepsilon^{-2\ell(2j-1)}\bigr),


  \sum_{j=1}^k \frac{1}{\sin^2{\frac{(2j-1)\pi}{4k+2}}}
  = \sum_{j=1}^k \left( (2k+1) +\sum_{\ell=1}^{2k}
    \bigl(2k+1-\ell)(\varepsilon^{2\ell(2j-1)} 
    +\varepsilon^{-2\ell(2j-1)}\bigr) \right) =


  = k(2k+1) + \sum_{\ell=1}^{2k} (2k+1-\ell) \sum_{j=1}^k 
  \bigl(\varepsilon^{2\ell(2j-1)} + \varepsilon^{-2\ell(2j-1)}\bigr).

Applying a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+b^{n-1}) again,


\sum_{j=1}^k \bigl(\varepsilon^{2\ell(2j-1)} + \varepsilon^{-2\ell(2j-1)}\bigr)
= -\varepsilon^{-2\ell(2k+1)}
+ \varepsilon^{-2\ell}\sum_{j=-k}^k \varepsilon^{4\ell j} =


= -\varepsilon^{-2\ell(2k+1)} + \varepsilon^{-2\ell}
\frac{\varepsilon^{2\ell(2k+1)}-\varepsilon^{-2\ell(2k+1)}}{
  \varepsilon^{2\ell}-\varepsilon^{-2\ell}}
= - (-1)^\ell + \varepsilon^{-2\ell}
\frac{(-1)^\ell-(-1)^\ell}{\varepsilon^{2\ell}-\varepsilon^{-2\ell}}
= (-1)^{\ell+1}.

Therefore,


\sum_{j=1}^k \frac{1}{\sin^2{\frac{(2j-1)\pi}{4k+2}}} 
= k(2k+1) - \sum_{\ell=1}^{2k} (2k+1-\ell) (-1)^\ell
= 2k(k+1).

Solution 3. We use U2k again and prove (1). The polynomial can be factorized as U_{2k}(x)=c\prod_{j=1}^{2k}(x-x_j), where c is the leading coefficient. Then


\frac{U_{2k}'(x)}{U_{2k}(x)} = \sum_{j=1}^{2k}\frac1{x-x_j},

and


\sum_{j=1}^{2k}\frac{1}{(x-x_j)^2} = 
\left(\sum_{j=1}^{2k}\frac{-1}{x-x_j}\right)'=
\left(-\frac{U_{2k}'(x)}{U_{2k}(x)}\right)' = 
-\frac{U_{2k}''(x)}{U_{2k}(x)}
+\left(\frac{U_{2k}'(x)}{U_{2k}(x)}\right)^2.

Substituting x=0 we obtain


\sum_{j=1}^{2k}\frac{1}{x_j^2} =
-\frac{U_{2k}''(0)}{U_{2k}(0)}+\left(\frac{U_{2k}'(0)}{U_{2k}(0)}\right)^2.
(2)

We compute the values U2k(0), U2k'(0), U2k''(0) by differentiating

U2k(cos t)sin t=sin ((2k+1)t)(3a)

twice.

-U2k'(cos t)sin2t+U2k(cos t)cos t=(2k+1)cos ((2k+1)t),(3b)
U2k''(cos t)sin3t-3U2k'(cos t)sin tcos t-U2k(cos t)sin t=-(2k+1)2sin ((2k+1)t).(3c)

By substituting t=\pi/2 in (3a--3c) we get U2k(0)=(-1)k and U2k'(0)=0, furthermore

U2k''(0)=(-1)k-(-1)k.(2k+1)2=(-1)k+1.4k(k+1).

Plugging these into (2),


\sum_{j=1}^{2k}\frac{1}{x_j^2} =
-\frac{(-1)^{k+1}\cdot4k(k+1)}{(-1)^k}+0^2 = 4k(k+1).


Statistics:

7 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Kovács 444 Áron, Mester Márton, Omer Cerrahoglu, Szabó 928 Attila.

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