(Ny side: Problem7 ----) |
|||
| Linje 1: | Linje 1: | ||
Problem7 | 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>. | ||
Revisjonen fra 4. feb. 2011 kl. 00:06
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>.
