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| (7 mellomliggende sideversjoner av samme bruker vises ikke) |
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| Problem5
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| Define <tex>d_n</tex> for natural <tex>n \geq 1</tex> as the determinant <tex>\underbrace{\begin{vmatrix}a&b&\ldots&0\\ c & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix}}_{\mbox{n}}</tex>. We have that <tex>d_{n+2} = \begin{vmatrix}a&b&\ldots&0\\ c & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix} = a \underbrace{\begin{vmatrix}a&b&\ldots&0\\ c & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix}}_{\mbox{=d_{n-1}}}-b\underbrace{\begin{vmatrix}c&b&\ldots&0\\ 0 & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix}}_{\mbox{n-1}} = ad_{n-1} -bc\underbrace{\begin{vmatrix}a&b&\ldots&0\\ c & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix}}_{\mbox{=d_{n-2}}} +b^2\underbrace{\begin{vmatrix}0&b&\ldots&0\\ 0 & a & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0&\ldots&a\end{vmatrix}}_{\mbox{=0}} =ad_{n-1}-bcd_{n-2}</tex>.
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| The characteristic polynomial for this difference equation is <tex>r^2-ar+bc</tex>, and solving it for zero yields <tex>r_1 = \frac{a+\sqrt{a^2-4bc}}{2}</tex>, <tex>r_2 = \frac{a-\sqrt{a^2-4bc}}{2}</tex>. Hereby assuming <tex>a^2 \not = 4bc</tex> (so <tex>r_1 \not = r_2</tex> and they are neither 0), we have <tex>d_1 = a</tex>, and <tex>d_2 = a^2-bc</tex>, so we can extend the definition of the <tex>d_n</tex>'s by setting <tex>d_0 = 1</tex>. Now, <tex>d_n =Ar_1^n+Br_2^n</tex> for constants <tex>A</tex> and <tex>B</tex>. But <tex>A+B = d_0 = 1</tex>, and <tex>Ar_1+Br_2 = a</tex>, so <tex>A(r_1-r_2)+r_2=a \Rightarrow A = \frac{a-r_2}{r_1-r_2}=\frac{r_1}{\sqrt{a^2-bc}}</tex>, and <tex>B = \frac{r_1-r_2-r_1}{r_1-r_2}= -\frac{r_2}{\sqrt{a^2-bc}}</tex>. Hence <tex>d_n = \frac{r_1^{n+1}-r_2^{n+1}}{\sqrt{a^2-bc}}</tex>, which for <tex>n \geq 1</tex> can be factorized to <tex>\frac{(r_1-r_2)(r_1^n+r_1^{n-1}r_2+...+r_1r_2^{n-1}+r_2^n)}{\sqrt{a^2-4bc}} = r_1^n+r_1^{n-1}r_2+...+r_1r_2^{n-1}+r_2^n</tex>.
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| We assumed in our calculations that <tex>a^2 \not = 4bc</tex>, but considering the fact that the determinant is continuous as a function in the variable <tex>a</tex> (keeping b and c constant) over (for simplicity) the complex numbers, and that <tex>d_n</tex> (considered a function in <tex>a</tex> on the set of complex points such that <tex>a^2 \not = bc</tex> which consists of all but maximally two points) in its last stated form have a unique continuous extension (the limits of <tex>d_n</tex> as <tex>a \to \pm 2 \sqrt{bc}</tex> both trivially exists), we know that this extension must coincide with the determinant for every complex point <tex>a</tex>.
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| Now, let <tex>f(n) = \prod^n_{k=1}(a-2\sqrt{bc}\cos(\frac{\pi k}{n+1})</tex>. Choose any points <tex>b</tex> and <tex>c</tex>, and consider <tex>f(n)</tex> as a polynomial in <tex>a</tex>. <tex>f(n)</tex> is the of degree <tex>n</tex>. <tex>d_n</tex> (being the determinant) is a polynomial in <tex>a</tex> (with the same choice of <tex>b</tex> and <tex>c</tex>), and is also of degree <tex>n</tex>. If we can show that <tex>d_n</tex> and <tex>f(n)</tex> share the same set of non-equal zeros, they must be equal up to a constant. This constant must in that case be 1, since <tex>f(1)=a=d_1</tex>.
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| To show this, let <tex>a = 2\sqrt{bc}\cos(\frac{\pi k}{n+1})</tex> for any integer <tex>k</tex> between <tex>1</tex> and <tex>n</tex>. Then <tex>r_1 = \frac{2\sqrt{bc}\cos(\frac{\pi k}{n+1})+\sqrt{4bc\cos^2(\frac{\pi k}{n+1})-4bc}}{2}=\sqrt{bc}\cos(\frac{\pi k}{n+1})+i\sqrt{bc}\sin(\frac{\pi k}{n+1}) = \sqrt{bc}e^{i\frac{\pi k}{n+1}}</tex>, and <tex>r_2 = \frac{2\sqrt{bc}\cos(\frac{\pi k}{n+1})-\sqrt{4bc\cos^2(\frac{\pi k}{n+1})-4bc}}{2}=\sqrt{bc}\cos(\frac{\pi k}{n+1})-i\sqrt{bc}\sin(\frac{\pi k}{n+1}) = \sqrt{bc}e^{-i\frac{\pi k}{n+1}}</tex>.
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| Hence
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| <tex>d_n =r_1^n+r_1^{n-1}r_2+...+r_1r_2^{n-1}+r_2^n = \sqrt{bc}^ne^{i\frac{n\pi k}{n+1}}+\sqrt{bc}^{n-1}e^{i\frac{(n-1)\pi k}{n+1}}\sqrt{bc}e^{-i\frac{\pi k}{n+1}} + ... +\sqrt{bc}^ne^{-in\frac{\pi k}{n+1}} \\ = \sqrt{bc}^ne^{-in\frac{\pi k}{n+1}}(e^{i\frac{n 2\pi k}{n+1}}+e^{i\frac{(n-1)2\pi k}{n+1}}+...+1) =\sqrt{bc}^ne^{-in\frac{\pi k}{n+1}}\frac{e^{i\frac{(n+1) 2\pi k}{n+1}} \ -1}{e^{i\frac{2\pi k}{n+1}}-1}=0</tex>
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| and we are done.
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