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Bruker:Karl Erik/Problem7 - Sideversjonshistorikk
2026-04-09T02:29:22Z
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Karl Erik: Erstatter siden med «Problem7
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2011-02-04T16:32:33Z
<p>Erstatter siden med «Problem7 ----»</p>
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<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. 16:32</td>
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<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>Problem7</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>Problem7</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;"><div>----</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>----</div></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;">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>.</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>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>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr>
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Karl Erik
https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem7&diff=3616&oldid=prev
Jarle på 4. feb. 2011 kl. 00:06
2011-02-04T00:06:25Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 00:06</td>
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<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>Problem7</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>Problem7</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;"><div>----</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>----</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></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><ins style="font-weight: bold; text-decoration: none;">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>.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></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><ins style="font-weight: bold; text-decoration: none;">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>.</ins></div></td></tr>
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Jarle
https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem7&diff=3607&oldid=prev
Karl Erik: Ny side: Problem7 ----
2011-02-03T22:17:57Z
<p>Ny side: Problem7 ----</p>
<p><b>Ny side</b></p><div>Problem7<br />
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Karl Erik