Matematikk.net - Brukerbidrag [nb]

Bruker:Karl Erik/Problem5

2011-02-04T20:32:09Z

Jarle: Tømmer siden

Bruker:Karl Erik/Problem10

2011-02-04T18:48:36Z

Jarle: Tømmer siden

Bruker:Karl Erik/Problem5

2011-02-04T03:27:44Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:25:46Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:24:21Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:22:52Z

Jarle:

Problem5
----
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>.

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>.

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 (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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:20:38Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:19:05Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:16:05Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:14:55Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem5

2011-02-04T03:13:44Z

Jarle:

Problem5
----
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>.

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>.

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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem10

2011-02-04T02:29:11Z

Jarle:

Problem10
----
Define <tex>f(n) = \sum^n_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n}</tex>. We will be using that <tex>\sum^m_{k=1}\cos \frac{2 \pi k}{m} = \Re(\sum^m_{k=1}e^{i\frac{2 \pi k}{m}}) = \Re(e^{i\frac{2 \pi}{m}}\frac{e^{mi\frac{2 \pi}{m}}-1}{e^{i\frac{2 \pi}{m}-1}}) = 0</tex>.

Consider a prime p. Then <tex>f(p) = \sum^p_{k=1} \gcd(k,p)\cos \frac{2 \pi k}{p} = \sum^p_{k=1} \cos \frac{2 \pi k}{p} +(p-1)\cos \frac{2 \pi p}{p} = p-1</tex>. We will show that if <tex>a_1,a_2,...,a_r</tex> are distinct prime factors, then <tex>f(a_1...a_r) = (a_1-1)...(a_r-1)</tex> by induction on <tex>r</tex>. We have already shown this for <tex>r = 1</tex>, so suppose it is true for <tex>s \leq r</tex>.

Let <tex>n = a_1...a_s</tex>, and let <tex>a_{s+1}</tex> be a prime not among the prime factors of <tex>n</tex>. <tex>f(na_{s+1}) = \sum^{na_{s+1}}_{k=1} \gcd(k,na_{s+1})\cos \frac{2 \pi k}{na_{s+1}} = \sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} + (a_{s+1}-1)\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n} \\ =\sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} +(a_{s+1}-1)f(n)</tex>.

So it remains to show that <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} = 0</tex> whenever <tex>p</tex> is a prime not among the distinct prime factors of <tex>n</tex>. Now, <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} =
\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{2n}_{k=n+1} \gcd(k,n)\cos \frac{2 \pi k}{np} + ... + \sum^{np}_{k=n(p-1)+1} \gcd(k,n)\cos \frac{2 \pi k}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{n}_{k=1} \gcd(k+n,n)\cos \frac{2 \pi (k+n)}{np} + ... + \sum^{n}_{k=1} \gcd(k+(p-1)n,n)\cos \frac{2 \pi (k+(p-1)n)}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\left(\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) \right)</tex>.

But for each <tex>k</tex> between <tex>1</tex> and <tex>n</tex>, we have <tex>\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) = \Re(e^{ i\frac{2 \pi k}{np}} +e^{i (\frac{2 \pi k}{np}+\frac{2\pi}{p})} + e^{i (\frac{2 \pi k}{np}+\frac{2\pi2}{p})} + ... + e^{i(\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p})}) \\ = \Re(e^{ i\frac{2 \pi k}{np}}(1+e^{i \frac{2\pi}{p}} + e^{i\frac{2\pi2}{p}} + ... + e^{i\frac{2\pi(p-1)}{p})}) = 0</tex>.

Thus we conclude that <tex>f(na_{s+1})=(a_{s+1}-1)f(n)=(a_1-1)...(a_s-1)(a_{s+1}-1)</tex>, and we are done.

Now, <tex>2010 = 2 \times 3 \times 5 \times 67</tex> is a product of distinct prime factors, so <tex>\sum^{2010}_{k=1} \gcd(k,2010)\cos \frac{2 \pi k}{2010} = f(2010) = (2-1)(3-1)(5-1)(67-1)=528</tex>.

Bruker:Karl Erik/Problem5

2011-02-04T01:28:28Z

Jarle:

Problem5
----
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>.

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>.

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), 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>.

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>.

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>.

Hence

<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>

and we are done.

Bruker:Karl Erik/Problem7

2011-02-04T00:06:25Z

Jarle:

Problem7
----
Consider the function <tex>f: (0,1] \to \mathbb{R}</tex> defined by <tex>f(x) = \frac{\tan x}{x}</tex>. We will show that it is increasing on its domain. We have <tex>f^{\prime}(x) = \frac{(\tan^2x+1)x-\tan(x)}{x^2}</tex>, and this function is non-negative if we can show that <tex>(\tan^2x+1)x-\tan(x) \geq 0</tex> for all <tex>x \in (0,1)</tex>. To verify this, consider the function <tex>g: [0,1) \to \mathbb{R}</tex> defined by <tex>g(x) = (\tan^2(x)+1)x-\tan(x)</tex>. <tex>g(0) = 0</tex>, and <tex>g^{\prime}(x) = \tan^2(x)+1+x(2\tan(x)(\tan^2(x)+1))-(\tan^2(x)+1) = 2x\tan(x)(\tan^2(x)+1)</tex> which clearly is non-negative on <tex>[0,1)</tex>. By this we conclude that <tex>f</tex> is an increasing function. Hence for any <tex>x \in (0,1)</tex> we have <tex>\frac{\tan(x)}{x} \leq \tan(1)</tex>, or equivalently <tex>\tan(x) \leq x\tan(1)</tex>.

Define <tex>b_n = a_n-n</tex> for every <tex>n \in \mathbb{N}</tex>. We know that <tex>a_n \in (n,n+1)</tex>, so <tex>b_n \in (0,1)</tex>. By the preceding paragraph, we have <tex>\tan(1)(\pi b_n) \geq \tan(\pi b_n)=\tan(\pi (a_n+n)) = \tan(\pi a_n)=\frac{1}{a_n} = \frac{1}{b_n+n} \geq \frac{1}{n+1}</tex>, hence <tex>b_n \geq \frac{1}{(n+1)\tan(1)\pi}</tex>. It follows that the sum <tex>\sum^N_{n=0}a_n-n \geq \frac{1}{\tan(1)\pi} \sum^N_{n=0} \frac{1}{n+1}</tex> and thus diverges as <tex>N \to \infty</tex>.