Bruker:Karl Erik/Problem5 - Sideversjonshistorikk

https://matematikk.net/w/index.php?action=history&feed=atom&title=Bruker%3AKarl_Erik%2FProblem5 Bruker:Karl Erik/Problem5 - Sideversjonshistorikk 2026-04-09T00:47:33Z Versjonshistorikk for denne siden på wikien MediaWiki 1.42.3 https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3653&oldid=prev Jarle: Tømmer siden 2011-02-04T20:32:09Z

<p>Tømmer siden</p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 20:32</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Linje 1:</td> <td colspan="2" class="diff-lineno">Linje 1:</td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Problem5</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">----</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">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>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">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> (choosing any <tex>b</tex> and <tex>c</tex> and keeping them 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 = 4bc</tex> which consists of all but maximally two points) in its last stated form (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">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> has 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>. </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">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>. </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Hence </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"><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></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">and we are done.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3627&oldid=prev Jarle på 4. feb. 2011 kl. 03:27 2011-02-04T03:27:44Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:27</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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> (choosing any <tex>b</tex> and <tex>c</tex> and keeping them 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 = <del style="font-weight: bold; text-decoration: none;">bc</del></tex> which consists of all but maximally two points) in its last stated form (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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> (choosing any <tex>b</tex> and <tex>c</tex> and keeping them 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 = <ins style="font-weight: bold; text-decoration: none;">4bc</ins></tex> which consists of all but maximally two points) in its last stated form (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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> has 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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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> has 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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3626&oldid=prev Jarle på 4. feb. 2011 kl. 03:25 2011-02-04T03:25:46Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:25</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7">Linje 7:</td> <td colspan="2" class="diff-lineno">Linje 7:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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> (choosing any <tex>b</tex> and <tex>c</tex> and keeping them 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 (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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> (choosing any <tex>b</tex> and <tex>c</tex> and keeping them 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 (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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> <del style="font-weight: bold; text-decoration: none;">is the of </del>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>. </div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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> <ins style="font-weight: bold; text-decoration: none;">has </ins>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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3625&oldid=prev Jarle på 4. feb. 2011 kl. 03:24 2011-02-04T03:24:21Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:24</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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> (<del style="font-weight: bold; text-decoration: none;">keeping </del>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 (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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> (<ins style="font-weight: bold; text-decoration: none;">choosing any <tex></ins>b<ins style="font-weight: bold; text-decoration: none;"></tex> </ins>and <ins style="font-weight: bold; text-decoration: none;"><tex></ins>c<ins style="font-weight: bold; text-decoration: none;"></tex> and keeping them </ins>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 (which is continuous) 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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3624&oldid=prev Jarle på 4. feb. 2011 kl. 03:22 2011-02-04T03:22:52Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:22</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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 <ins style="font-weight: bold; text-decoration: none;">(which is continuous) </ins>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>. Thus we consider <tex>d_n</tex> extended, and therefore equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3623&oldid=prev Jarle på 4. feb. 2011 kl. 03:20 2011-02-04T03:20:38Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:20</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 that <tex>r_1 \not = r_2</tex>, neither are 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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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>. Thus we consider <tex>d_n</tex> extended, and equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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>. Thus we consider <tex>d_n</tex> extended, and <ins style="font-weight: bold; text-decoration: none;">therefore </ins>equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3622&oldid=prev Jarle på 4. feb. 2011 kl. 03:19 2011-02-04T03:19:05Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:19</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l3">Linje 3:</td> <td colspan="2" class="diff-lineno">Linje 3:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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> <del style="font-weight: bold; text-decoration: none;">and they </del>are <del style="font-weight: bold; text-decoration: none;">neither </del>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>. </div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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 <ins style="font-weight: bold; text-decoration: none;">that </ins><tex>r_1 \not = r_2</tex><ins style="font-weight: bold; text-decoration: none;">, neither </ins>are 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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. Thus we consider <tex>d_n</tex> extended, and equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. Thus we consider <tex>d_n</tex> extended, and equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</tex>.</div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3621&oldid=prev Jarle på 4. feb. 2011 kl. 03:16 2011-02-04T03:16:05Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:16</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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<ins style="font-weight: bold; text-decoration: none;"></tex>. Thus we consider <tex>d_n</tex> extended, and equal to the determinant for all complex numbers <tex>a</tex>, <tex>b</tex> and <tex>c</ins></tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3620&oldid=prev Jarle på 4. feb. 2011 kl. 03:14 2011-02-04T03:14:55Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:14</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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 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>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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 <ins style="font-weight: bold; text-decoration: none;">maximally </ins>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>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem5&diff=3619&oldid=prev Jarle på 4. feb. 2011 kl. 03:13 2011-02-04T03:13:44Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 03:13</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5">Linje 5:</td> <td colspan="2" class="diff-lineno">Linje 5:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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><del style="font-weight: bold; text-decoration: none;">, </del>(keeping b and c constant), and that <tex>d_n</tex> (considered a function in <tex>a</tex> on the set of points such that <tex>a^2 \not = bc</tex> which consists of all but 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 point <tex>a</tex>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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) <ins style="font-weight: bold; text-decoration: none;">over (for simplicity) the complex numbers</ins>, and that <tex>d_n</tex> (considered a function in <tex>a</tex> on the set of <ins style="font-weight: bold; text-decoration: none;">complex </ins>points such that <tex>a^2 \not = bc</tex> which consists of all but 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 <ins style="font-weight: bold; text-decoration: none;">complex </ins>point <tex>a</tex>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>. </div></td></tr> </table>

Jarle